100 results found for "Math Introduction To Trigonometry Trigonometric Ratios Of An Acute Angle In A Right Angled Triangle" in Class 10.
एक समकोण त्रिभुज का आधार (x+5), ऊंचाई (x+9) और क्षेत्रफल (95) है। सही समीकरण कौन-सा है?
A right triangle has base (x+5), height (x+9), and area (95). Which equation is correct?
#quadratic-equations
#word-problem
#triangle-area
#expert
A \(x^2+14x-145=0\)
B \(x^2+14x-95=0\)
C \(x^2+14x+45=0\)
D \(x^2+14x-190=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+14x-145=0\)
Step 1
Concept
The area is (\frac{1}{2}(x+5)(x+9)=95). Thus ((x+5)(x+9)=190) and \(x^2+14x-145=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2+14x-145=0\). The area is (\frac{1}{2}(x+5)(x+9)=95). Thus ((x+5)(x+9)=190) and \(x^2+14x-145=0\).
Step 3
Exam Tip
क्षेत्रफल (\frac{1}{2}(x+5)(x+9)=95) होगा। इसलिए ((x+5)(x+9)=190) और \(x^2+14x-145=0\)।
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एक समकोण त्रिभुज का आधार (x+4), ऊंचाई (x+8) और क्षेत्रफल (72) है। सही समीकरण कौन-सा है?
A right triangle has base (x+4), height (x+8), and area (72). Which equation is correct?
#quadratic-equations
#word-problem
#triangle-area
#expert
A \(x^2+12x-112=0\)
B \(x^2+12x-72=0\)
C \(x^2+12x+32=0\)
D \(x^2+12x-144=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+12x-112=0\)
Step 1
Concept
The area is (\frac{1}{2}(x+4)(x+8)=72). Thus ((x+4)(x+8)=144) and \(x^2+12x-112=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2+12x-112=0\). The area is (\frac{1}{2}(x+4)(x+8)=72). Thus ((x+4)(x+8)=144) and \(x^2+12x-112=0\).
Step 3
Exam Tip
क्षेत्रफल (\frac{1}{2}(x+4)(x+8)=72) होगा। इसलिए ((x+4)(x+8)=144) और \(x^2+12x-112=0\)।
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एक समकोण त्रिभुज का आधार (x+3), ऊंचाई (x+7) और क्षेत्रफल (55) है। सही समीकरण कौन-सा है?
A right triangle has base (x+3), height (x+7), and area (55). Which equation is correct?
#quadratic-equations
#word-problem
#triangle-area
#expert
A \(x^2+10x-89=0\)
B \(x^2+10x-55=0\)
C \(x^2+10x+21=0\)
D \(x^2+10x-110=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+10x-89=0\)
Step 1
Concept
The area is (\frac{1}{2}(x+3)(x+7)=55). Thus ((x+3)(x+7)=110) and \(x^2+10x-89=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2+10x-89=0\). The area is (\frac{1}{2}(x+3)(x+7)=55). Thus ((x+3)(x+7)=110) and \(x^2+10x-89=0\).
Step 3
Exam Tip
क्षेत्रफल (\frac{1}{2}(x+3)(x+7)=55) होगा। इसलिए ((x+3)(x+7)=110) और \(x^2+10x-89=0\)।
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एक समकोण त्रिभुज का आधार (x+2), ऊंचाई (x+6) और क्षेत्रफल (40) है। सही समीकरण कौन-सा है?
A right triangle has base (x+2), height (x+6), and area (40). Which equation is correct?
#quadratic-equations
#word-problem
#triangle-area
#hard
A \(x^2+8x-68=0\)
B \(x^2+8x-40=0\)
C \(x^2+8x+12=0\)
D \(x^2+8x-80=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+8x-68=0\)
Step 1
Concept
The area is (\frac{1}{2}(x+2)(x+6)=40). Thus ((x+2)(x+6)=80) and \(x^2+8x-68=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2+8x-68=0\). The area is (\frac{1}{2}(x+2)(x+6)=40). Thus ((x+2)(x+6)=80) and \(x^2+8x-68=0\).
Step 3
Exam Tip
क्षेत्रफल (\frac{1}{2}(x+2)(x+6)=40) होगा। इसलिए ((x+2)(x+6)=80) और \(x^2+8x-68=0\)।
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एक समकोण त्रिभुज का आधार (x+1), ऊंचाई (x+3) और क्षेत्रफल (30) है। सही समीकरण कौन-सा है?
A right triangle has base (x+1), height (x+3), and area (30). Which equation is correct?
#quadratic-equations
#word-problem
#triangle-area
#hard
A \(x^2+4x-57=0\)
B \(x^2+4x-30=0\)
C \(x^2+4x+3=0\)
D \(x^2+4x-60=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+4x-57=0\)
Step 1
Concept
The area is (\frac{1}{2}(x+1)(x+3)=30). Thus ((x+1)(x+3)=60) and \(x^2+4x-57=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2+4x-57=0\). The area is (\frac{1}{2}(x+1)(x+3)=30). Thus ((x+1)(x+3)=60) and \(x^2+4x-57=0\).
Step 3
Exam Tip
क्षेत्रफल (\frac{1}{2}(x+1)(x+3)=30) होगा। इसलिए ((x+1)(x+3)=60) और \(x^2+4x-57=0\)।
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एक समकोण त्रिभुज का आधार (x), ऊँचाई (x+2) और क्षेत्रफल (24) है। सही द्विघात समीकरण कौन-सा है?
A right triangle has base (x), height (x+2), and area (24). Which quadratic equation is correct?
#quadratic-equations
#word-problem
#triangle-area
#medium
A \(x^2+2x-48=0\)
B \(x^2+2x-24=0\)
C \(2x^2+2x-24=0\)
D \(x^2-2x-48=0\)
Explanation opens after your attempt
Correct Answer
A. \(x^2+2x-48=0\)
Step 1
Concept
The area is (\frac{1}{2}x(x+2)=24). Thus (x(x+2)=48), giving \(x^2+2x-48=0\).
Step 2
Why this answer is correct
The correct answer is A. \(x^2+2x-48=0\). The area is (\frac{1}{2}x(x+2)=24). Thus (x(x+2)=48), giving \(x^2+2x-48=0\).
Step 3
Exam Tip
क्षेत्रफल (\frac{1}{2}x(x+2)=24) होगा। इसलिए (x(x+2)=48) और \(x^2+2x-48=0\) मिलता है।
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एक समकोण त्रिभुज का क्षेत्रफल (84) वर्ग सेमी है और एक लम्ब पक्ष दूसरे से (5) सेमी अधिक है। छोटे लम्ब पक्ष की लंबाई क्या है?
The area of a right triangle is (84) square cm and one perpendicular side is (5) cm more than the other. What is the shorter perpendicular side?
#quadratic equations
#right triangle
#area
A (7) सेमी / (7) cm
B (8) सेमी / (8) cm
C (12) सेमी / (12) cm
D (14) सेमी / (14) cm
Explanation opens after your attempt
Correct Answer
C. (12) सेमी / (12) cm
Step 1
Concept
Let the shorter side be (x). Then (\frac{1}{2}x(x+5)=84). This gives \(x^2+5x-168=0\), so (x=12).
Step 2
Why this answer is correct
The correct answer is C. (12) सेमी / (12) cm. Let the shorter side be (x). Then (\frac{1}{2}x(x+5)=84). This gives \(x^2+5x-168=0\), so (x=12).
Step 3
Exam Tip
छोटा पक्ष (x) हो तो (\frac{1}{2}x(x+5)=84)। इससे \(x^2+5x-168=0\) और (x=12) है।
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एक समकोण त्रिभुज के दो छोटे पक्ष (x) सेमी और (x+2) सेमी हैं तथा कर्ण (10) सेमी है। बड़ा छोटा पक्ष क्या है?
The two shorter sides of a right triangle are (x) cm and (x+2) cm, and the hypotenuse is (10) cm. What is the larger shorter side?
#quadratic equations
#right triangle
#application
A (6) सेमी / (6) cm
B (7) सेमी / (7) cm
C (8) सेमी / (8) cm
D (9) सेमी / (9) cm
Explanation opens after your attempt
Correct Answer
C. (8) सेमी / (8) cm
Step 1
Concept
We get (x-2 +(x+2)2 =100). This gives (x=6), so the larger shorter side is (8) cm.
Step 2
Why this answer is correct
The correct answer is C. (8) सेमी / (8) cm. We get (x-2 +(x+2)2 =100). This gives (x=6), so the larger shorter side is (8) cm.
Step 3
Exam Tip
(x-2 +(x+2)2 =100) बनता है। इससे (x=6) मिलता है इसलिए बड़ा छोटा पक्ष (8) सेमी है।
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एक समकोण त्रिभुज में लम्ब आधार से (7) सेमी अधिक है और कर्ण (25) सेमी है। आधार कितना है?
In a right triangle, the perpendicular is (7) cm more than the base and the hypotenuse is (25) cm. What is the base?
#quadratic equations
#right triangle
#pythagoras
A (15) सेमी / (15) cm
B (16) सेमी / (16) cm
C (18) सेमी / (18) cm
D (20) सेमी / (20) cm
Explanation opens after your attempt
Correct Answer
A. (15) सेमी / (15) cm
Step 1
Concept
The base is (x) and perpendicular is (x+7). From (x-2 +(x+7)2 =252 ), (x=15).
Step 2
Why this answer is correct
The correct answer is A. (15) सेमी / (15) cm. The base is (x) and perpendicular is (x+7). From (x-2 +(x+7)2 =252 ), (x=15).
Step 3
Exam Tip
आधार (x) और लम्ब (x+7) है। (x-2 +(x+7)2 =252 ) से (x=15) मिलता है।
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एक समकोण त्रिभुज का कर्ण (17,सेमी) है और एक भुजा दूसरी से (7,सेमी) अधिक है। छोटी भुजा क्या है?
The hypotenuse of a right triangle is (17,cm) and one side is (7,cm) more than the other. What is the smaller side?
#quadratic equations
#right triangle
#pythagoras
A (6,सेमी) / (6,cm)
B (7,सेमी) / (7,cm)
C (8,सेमी) / (8,cm)
D (9,सेमी) / (9,cm)
Explanation opens after your attempt
Correct Answer
C. (8,सेमी) / (8,cm)
Step 1
Concept
If the smaller side is (x), then (x-2 +(x+7)2 =172 ). This gives (x=8).
Step 2
Why this answer is correct
The correct answer is C. (8,सेमी) / (8,cm\(). If the smaller side is (x), then (x^2+(x+7)^2=17^2). This gives (x=8).\)
Step 3
Exam Tip
छोटी भुजा (x) हो तो (x-2 +(x+7)2 =172 )। इससे (x=8) मिलता है।
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एक समकोण त्रिभुज में छोटी भुजा (x cm), दूसरी भुजा (x+7 cm) और कर्ण (x+8 cm) है। छोटी भुजा क्या है?
In a right triangle, the shorter side is (x cm), the other side is (x+7 cm), and the hypotenuse is (x+8 cm). What is the shorter side?
#quadratic equations
#right triangle
#pythagoras
A (5 cm)
B (6 cm)
C (7 cm)
D (8 cm)
Explanation opens after your attempt
Step 1
Concept
By Pythagoras, (x-2 +(x+7)2 =(x+8)2 ). This gives \(x^2-2x-15=0\), so (x=5).
Step 2
Why this answer is correct
The correct answer is A. (5 cm\(). By Pythagoras, (x^2+(x+7)^2=(x+8)^2). This gives (x^2-2x-15=0), so (x=5).\)
Step 3
Exam Tip
पाइथागोरस से (x-2 +(x+7)2 =(x+8)2 )। इससे \(x^2-2x-15=0\), इसलिए (x=5)।
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एक समकोण त्रिभुज में कर्ण (185) सेमी है और एक लम्ब भुजा दूसरी से (37) सेमी अधिक है। छोटी लम्ब भुजा क्या है?
In a right triangle, the hypotenuse is (185) cm and one leg is (37) cm more than the other. What is the smaller leg?
#quadratic equations
#right triangle
#pythagoras
A (96) सेमी / (96) cm
B (111) सेमी / (111) cm
C (120) सेमी / (120) cm
D (148) सेमी / (148) cm
Explanation opens after your attempt
Correct Answer
B. (111) सेमी / (111) cm
Step 1
Concept
If the smaller leg is (x), then (x-2 +(x+37)2 =1852 ). This gives (x=111).
Step 2
Why this answer is correct
The correct answer is B. (111) सेमी / (111) cm. If the smaller leg is (x), then (x-2 +(x+37)2 =1852 ). This gives (x=111).
Step 3
Exam Tip
छोटी भुजा (x) हो तो (x-2 +(x+37)2 =1852 )। इससे (x=111) मिलता है।
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एक समकोण त्रिभुज की लम्ब भुजाएँ (x) सेमी और (x+30) सेमी हैं तथा कर्ण (150) सेमी है। छोटी लम्ब भुजा क्या है?
The legs of a right triangle are (x) cm and (x+30) cm and the hypotenuse is (150) cm. What is the smaller leg?
#quadratic equations
#pythagoras
#right triangle
A (75) सेमी / (75) cm
B (80) सेमी / (80) cm
C (90) सेमी / (90) cm
D (120) सेमी / (120) cm
Explanation opens after your attempt
Correct Answer
C. (90) सेमी / (90) cm
Step 1
Concept
By Pythagoras, (x-2 +(x+30)2 =1502 ), giving (x=90). Form the equation by taking the hypotenuse as the largest side.
Step 2
Why this answer is correct
The correct answer is C. (90) सेमी / (90) cm. By Pythagoras, (x-2 +(x+30)2 =1502 ), giving (x=90). Form the equation by taking the hypotenuse as the largest side.
Step 3
Exam Tip
पाइथागोरस से (x-2 +(x+30)2 =1502 ), जिससे (x=90) है। कर्ण को सबसे बड़ी भुजा मानकर समीकरण बनाएं।
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एक समकोण त्रिभुज में कर्ण (145) सेमी है और एक लम्ब भुजा दूसरी से (17) सेमी अधिक है। छोटी लम्ब भुजा क्या है?
In a right triangle, the hypotenuse is (145) cm and one leg is (17) cm more than the other. What is the smaller leg?
#quadratic equations
#right triangle
#pythagoras
A (84) सेमी / (84) cm
B (88) सेमी / (88) cm
C (95) सेमी / (95) cm
D (112) सेमी / (112) cm
Explanation opens after your attempt
Correct Answer
A. (84) सेमी / (84) cm
Step 1
Concept
If the smaller leg is (x), then (x-2 +(x+17)2 =1452 ). This gives (x=84).
Step 2
Why this answer is correct
The correct answer is A. (84) सेमी / (84) cm. If the smaller leg is (x), then (x-2 +(x+17)2 =1452 ). This gives (x=84).
Step 3
Exam Tip
छोटी भुजा (x) हो तो (x-2 +(x+17)2 =1452 )। इससे (x=84) मिलता है।
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एक समकोण त्रिभुज की लम्ब भुजाएँ (x) सेमी और (x+24) सेमी हैं तथा कर्ण (120) सेमी है। छोटी लम्ब भुजा क्या है?
The legs of a right triangle are (x) cm and (x+24) cm and the hypotenuse is (120) cm. What is the smaller leg?
#quadratic equations
#pythagoras
#right triangle
A (66) सेमी / (66) cm
B (72) सेमी / (72) cm
C (80) सेमी / (80) cm
D (96) सेमी / (96) cm
Explanation opens after your attempt
Correct Answer
B. (72) सेमी / (72) cm
Step 1
Concept
By Pythagoras, (x-2 +(x+24)2 =1202 ), giving (x=72). Form the equation by taking the hypotenuse as the largest side.
Step 2
Why this answer is correct
The correct answer is B. (72) सेमी / (72) cm. By Pythagoras, (x-2 +(x+24)2 =1202 ), giving (x=72). Form the equation by taking the hypotenuse as the largest side.
Step 3
Exam Tip
पाइथागोरस से (x-2 +(x+24)2 =1202 ), जिससे (x=72) है। कर्ण को सबसे बड़ी भुजा मानकर समीकरण बनाएं।
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एक समकोण त्रिभुज में कर्ण (109) सेमी है और एक लम्ब भुजा दूसरी से (19) सेमी अधिक है। छोटी लम्ब भुजा क्या है?
In a right triangle, the hypotenuse is (109) cm and one leg is (19) cm more than the other. What is the smaller leg?
#quadratic equations
#right triangle
#pythagoras
A (60) सेमी / (60) cm
B (65) सेमी / (65) cm
C (72) सेमी / (72) cm
D (84) सेमी / (84) cm
Explanation opens after your attempt
Correct Answer
A. (60) सेमी / (60) cm
Step 1
Concept
If the smaller leg is (x), then (x-2 +(x+19)2 =1092 ). This gives (x=60).
Step 2
Why this answer is correct
The correct answer is A. (60) सेमी / (60) cm. If the smaller leg is (x), then (x-2 +(x+19)2 =1092 ). This gives (x=60).
Step 3
Exam Tip
छोटी भुजा (x) हो तो (x-2 +(x+19)2 =1092 )। इससे (x=60) मिलता है।
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एक समकोण त्रिभुज की लम्ब भुजाएँ (x) सेमी और (x+18) सेमी हैं तथा कर्ण (90) सेमी है। छोटी लम्ब भुजा क्या है?
The legs of a right triangle are (x) cm and (x+18) cm and the hypotenuse is (90) cm. What is the smaller leg?
#quadratic equations
#pythagoras
#right triangle
A (48) सेमी / (48) cm
B (54) सेमी / (54) cm
C (60) सेमी / (60) cm
D (72) सेमी / (72) cm
Explanation opens after your attempt
Correct Answer
B. (54) सेमी / (54) cm
Step 1
Concept
By Pythagoras, (x-2 +(x+18)2 =902 ), giving (x=54). Form the equation by taking the hypotenuse as the largest side.
Step 2
Why this answer is correct
The correct answer is B. (54) सेमी / (54) cm. By Pythagoras, (x-2 +(x+18)2 =902 ), giving (x=54). Form the equation by taking the hypotenuse as the largest side.
Step 3
Exam Tip
पाइथागोरस से (x-2 +(x+18)2 =902 ), जिससे (x=54) है। कर्ण को सबसे बड़ी भुजा मानकर समीकरण बनाएं।
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एक समकोण त्रिभुज में कर्ण (41) है और एक भुजा दूसरी से (31) अधिक है। छोटी भुजा क्या है?
In a right triangle, the hypotenuse is (41) and one leg is (31) more than the other. What is the smaller leg?
#quadratic-equations
#word-problems
#right-triangle
A (9)
B (40)
C (31)
D (10)
Explanation opens after your attempt
Step 1
Concept
If the smaller leg is (x), the other is (x+31). From (x-2 +(x+31)2 =412 ), (x=9).
Step 2
Why this answer is correct
The correct answer is A. (9). If the smaller leg is (x), the other is (x+31). From (x-2 +(x+31)2 =412 ), (x=9).
Step 3
Exam Tip
यदि छोटी भुजा (x) है, तो दूसरी (x+31) होगी। (x-2 +(x+31)2 =412 ) से (x=9) मिलता है।
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एक समकोण त्रिभुज में कर्ण (85) सेमी है और एक लम्ब भुजा दूसरी से (13) सेमी अधिक है। छोटी लम्ब भुजा क्या है?
In a right triangle, the hypotenuse is (85) cm and one leg is (13) cm more than the other. What is the smaller leg?
#quadratic equations
#right triangle
#pythagoras
A (48) सेमी / (48) cm
B (51) सेमी / (51) cm
C (60) सेमी / (60) cm
D (72) सेमी / (72) cm
Explanation opens after your attempt
Correct Answer
B. (51) सेमी / (51) cm
Step 1
Concept
If the smaller leg is (x), then (x-2 +(x+13)2 =852 ). This gives (x=51).
Step 2
Why this answer is correct
The correct answer is B. (51) सेमी / (51) cm. If the smaller leg is (x), then (x-2 +(x+13)2 =852 ). This gives (x=51).
Step 3
Exam Tip
छोटी भुजा (x) हो तो (x-2 +(x+13)2 =852 )। इससे (x=51) मिलता है।
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एक समकोण त्रिभुज की लम्ब भुजाएँ (x) सेमी और (x+14) सेमी हैं तथा कर्ण (70) सेमी है। छोटी लम्ब भुजा क्या है?
The legs of a right triangle are (x) cm and (x+14) cm and the hypotenuse is (70) cm. What is the smaller leg?
#quadratic equations
#pythagoras
#right triangle
A (42) सेमी / (42) cm
B (44) सेमी / (44) cm
C (46) सेमी / (46) cm
D (48) सेमी / (48) cm
Explanation opens after your attempt
Correct Answer
A. (42) सेमी / (42) cm
Step 1
Concept
By Pythagoras, (x-2 +(x+14)2 =702 ), giving (x=42). Form the equation by taking the hypotenuse as the largest side.
Step 2
Why this answer is correct
The correct answer is A. (42) सेमी / (42) cm. By Pythagoras, (x-2 +(x+14)2 =702 ), giving (x=42). Form the equation by taking the hypotenuse as the largest side.
Step 3
Exam Tip
पाइथागोरस से (x-2 +(x+14)2 =702 ), जिससे (x=42) है। कर्ण को सबसे बड़ी भुजा मानकर समीकरण बनाएं।
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एक समकोण त्रिभुज में कर्ण (34) है और एक भुजा दूसरी से (14) अधिक है। छोटी भुजा क्या है?
In a right triangle, the hypotenuse is (34) and one side is (14) more than the other. What is the smaller side?
#quadratic-equations
#word-problems
#right-triangle
A (16)
B (30)
C (18)
D (20)
Explanation opens after your attempt
Step 1
Concept
If the smaller side is (x), the other side is (x+14). From (x-2 +(x+14)2 =342 ), (x=16).
Step 2
Why this answer is correct
The correct answer is A. (16). If the smaller side is (x), the other side is (x+14). From (x-2 +(x+14)2 =342 ), (x=16).
Step 3
Exam Tip
यदि छोटी भुजा (x) है, तो दूसरी (x+14) होगी। (x-2 +(x+14)2 =342 ) से (x=16) मिलता है।
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एक समकोण त्रिभुज में कर्ण (17) है और एक भुजा दूसरी से (7) अधिक है। छोटी भुजा क्या है?
In a right triangle, the hypotenuse is (17) and one leg is (7) more than the other. What is the smaller leg?
#quadratic-equations
#word-problems
#right-triangle
A (8)
B (15)
C (10)
D (7)
Explanation opens after your attempt
Step 1
Concept
If the smaller leg is (x), the other is (x+7). From (x-2 +(x+7)2 =172 ), (x=8).
Step 2
Why this answer is correct
The correct answer is A. (8). If the smaller leg is (x), the other is (x+7). From (x-2 +(x+7)2 =172 ), (x=8).
Step 3
Exam Tip
यदि छोटी भुजा (x) है, तो दूसरी (x+7) होगी। (x-2 +(x+7)2 =172 ) से (x=8) मिलता है।
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एक समकोण त्रिभुज की लम्ब भुजाएँ (x) सेमी और (x+12) सेमी हैं तथा कर्ण (60) सेमी है। (x) का मान क्या है?
The legs of a right triangle are (x) cm and (x+12) cm, and the hypotenuse is (60) cm. What is the value of (x)?
#quadratic equations
#pythagoras
#right triangle
A (30)
B (36)
C (42)
D (48)
Explanation opens after your attempt
Step 1
Concept
By Pythagoras, (x-2 +(x+12)2 =602 ), giving (x=36). Always take the hypotenuse as the largest side.
Step 2
Why this answer is correct
The correct answer is B. (36). By Pythagoras, (x-2 +(x+12)2 =602 ), giving (x=36). Always take the hypotenuse as the largest side.
Step 3
Exam Tip
पाइथागोरस से (x-2 +(x+12)2 =602 ), जिससे (x=36) है। कर्ण को हमेशा सबसे बड़ी भुजा मानें।
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एक समकोण त्रिभुज की लम्ब भुजाएँ (x) सेमी और (x+8) सेमी हैं तथा कर्ण (40) सेमी है। (x) का मान क्या है?
The legs of a right triangle are (x) cm and (x+8) cm, and the hypotenuse is (40) cm. What is the value of (x)?
#quadratic equations
#pythagoras
#right triangle
A (20)
B (24)
C (28)
D (32)
Explanation opens after your attempt
Step 1
Concept
By Pythagoras, (x-2 +(x+8)2 =402 ), giving (x=24). In a right triangle, forming the correct equation is most important.
Step 2
Why this answer is correct
The correct answer is B. (24). By Pythagoras, (x-2 +(x+8)2 =402 ), giving (x=24). In a right triangle, forming the correct equation is most important.
Step 3
Exam Tip
पाइथागोरस से (x-2 +(x+8)2 =402 ), जिससे (x=24) है। समकोण त्रिभुज में सही समीकरण बनाना सबसे जरूरी है।
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एक समकोण त्रिभुज की लम्ब भुजाएँ (x) सेमी और (x+5) सेमी हैं तथा कर्ण (25) सेमी है। छोटी लम्ब भुजा क्या है?
The legs of a right triangle are (x) cm and (x+5) cm, and the hypotenuse is (25) cm. What is the smaller leg?
#quadratic equations
#pythagoras
#right triangle
A (12) सेमी / (12) cm
B (15) सेमी / (15) cm
C (18) सेमी / (18) cm
D (20) सेमी / (20) cm
Explanation opens after your attempt
Correct Answer
B. (15) सेमी / (15) cm
Step 1
Concept
By Pythagoras, (x-2 +(x+5)2 =252 ), giving (x=15). Always take the hypotenuse as the largest side.
Step 2
Why this answer is correct
The correct answer is B. (15) सेमी / (15) cm. By Pythagoras, (x-2 +(x+5)2 =252 ), giving (x=15). Always take the hypotenuse as the largest side.
Step 3
Exam Tip
पाइथागोरस से (x-2 +(x+5)2 =252 ), जिससे (x=15) है। कर्ण को हमेशा सबसे बड़ी भुजा मानें।
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एक समकोण त्रिभुज में लम्ब भुजाएँ (x) सेमी और (x+1) सेमी हैं तथा कर्ण (5) सेमी है। छोटी भुजा क्या है?
A right triangle has legs (x) cm and (x+1) cm and hypotenuse (5) cm. What is the smaller leg?
#quadratic equations
#pythagoras
#right triangle
A (2) सेमी / (2) cm
B (3) सेमी / (3) cm
C (4) सेमी / (4) cm
D (5) सेमी / (5) cm
Explanation opens after your attempt
Correct Answer
B. (3) सेमी / (3) cm
Step 1
Concept
(x-2 +(x+1)2 =52 ) gives (x=3). After solving, check the (3,4,5) triplet.
Step 2
Why this answer is correct
The correct answer is B. (3) सेमी / (3) cm. (x-2 +(x+1)2 =52 ) gives (x=3). After solving, check the (3,4,5) triplet.
Step 3
Exam Tip
(x-2 +(x+1)2 =52 ) से (x=3) मिलता है। हल के बाद (3,4,5) त्रिक की जाँच करें।
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एक त्रिभुज का क्षेत्रफल \(60,सेमी^2\) है। उसका आधार ऊँचाई से (4,सेमी) अधिक है। ऊँचाई क्या है?
The area of a triangle is (60,cm\(^2). Its base is (4\),cm) more than its height. What is the height?
#quadratic equations
#triangle area
#application
A (8,सेमी) / (8,cm)
B (10,सेमी) / (10,cm)
C (12,सेमी) / (12,cm)
D (14,सेमी) / (14,cm)
Explanation opens after your attempt
Correct Answer
B. (10,सेमी) / (10,cm)
Step 1
Concept
If height is (x), then (\frac{1}{2}x(x+4)=60). Hence (x=10).
Step 2
Why this answer is correct
The correct answer is B. (10,सेमी) / (10,cm\(). If height is (x), then (\frac{1}{2}x(x+4)=60). Hence (x=10).\)
Step 3
Exam Tip
यदि ऊँचाई (x) है तो (\frac{1}{2}x(x+4)=60)। इसलिए (x=10) है।
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एक त्रिभुज का आधार ऊँचाई से (6 cm) अधिक है और क्षेत्रफल \(140 cm^2\) है। ऊँचाई क्या है?
The base of a triangle is (6 cm) more than its height, and its area is (140 cm\(^2). What is the height\)?
#quadratic equations
#triangle area
#application
A (10 cm)
B (14 cm)
C (16 cm)
D (20 cm)
Explanation opens after your attempt
Correct Answer
B. (14 cm)
Step 1
Concept
Let height be (x), then (\frac{1}{2}x(x+6)=140). This gives \(x^2+6x-280=0\), so (x=14).
Step 2
Why this answer is correct
The correct answer is B. (14 cm\(). Let height be (x), then (\frac{1}{2}x(x+6)=140). This gives (x^2+6x-280=0), so (x=14).\)
Step 3
Exam Tip
ऊँचाई (x) हो, तो (\frac{1}{2}x(x+6)=140)। इससे \(x^2+6x-280=0\), इसलिए (x=14)।
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किसी पोस्टर में त्रिभुज का उपयोग खतरे या चेतावनी जैसा प्रभाव क्यों दे सकता है?
Why can use of a triangle in a poster give a danger or warning-like effect?
#triangle
#sharp angles
#poster
A क्योंकि उसमें नुकीले कोण होते हैं / Because it has sharp angles
B क्योंकि वह हमेशा नीला होता है / Because it is always blue
C क्योंकि वह छूने से खुरदरा होता है / Because it is rough to touch
D क्योंकि वह केवल ऋणात्मक स्थान है / Because it is only negative space
Explanation opens after your attempt
Correct Answer
A. क्योंकि उसमें नुकीले कोण होते हैं / Because it has sharp angles
Step 1
Concept
Sharp angles can suggest sharpness and caution. Exam tip: understand shape mood through context.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि उसमें नुकीले कोण होते हैं / Because it has sharp angles. Sharp angles can suggest sharpness and caution. Exam tip: understand shape mood through context.
Step 3
Exam Tip
नुकीले कोण तीखापन और सावधानी का संकेत दे सकते हैं। परीक्षा में shape mood को संदर्भ से समझें।
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वृत्त और त्रिभुज में समानता किस आधार पर है?
On what basis are circle and triangle similar?
#shape
#circle
#triangle
A दोनों द्वि आयामी आकार हैं / Both are two-dimensional shapes
B दोनों त्रि आयामी रूप हैं / Both are three-dimensional forms
C दोनों वास्तविक बनावट हैं / Both are actual textures
D दोनों रंग ताप हैं / Both are colour temperatures
Explanation opens after your attempt
Correct Answer
A. दोनों द्वि आयामी आकार हैं / Both are two-dimensional shapes
Step 1
Concept
Circle and triangle are both flat shapes. Exam tip: write flat closed area as shape.
Step 2
Why this answer is correct
The correct answer is A. दोनों द्वि आयामी आकार हैं / Both are two-dimensional shapes. Circle and triangle are both flat shapes. Exam tip: write flat closed area as shape.
Step 3
Exam Tip
वृत्त और त्रिभुज दोनों समतल आकार हैं। परीक्षा में flat closed area को shape लिखें।
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संख्या रेखा पर \( \sqrt{5} \) बनाने के लिए (2) और (1) लंबाई वाली समकोण भुजाओं का कर्ण क्या होगा?
To construct \( \sqrt{5} \) on the number line, what is the hypotenuse of a right triangle with perpendicular sides (2) and (1)?
#number-line
#geometric-construction
#pythagoras
A (3)
B \( \sqrt{3} \)
C \( \sqrt{5} \)
D (5)
Explanation opens after your attempt
Correct Answer
C. \( \sqrt{5} \)
Step 1
Concept
By Pythagoras, the hypotenuse is \( \sqrt{2^2+1^2}=\sqrt{5} \). Right triangles help in square-root construction.
Step 2
Why this answer is correct
The correct answer is C. \( \sqrt{5} \). By Pythagoras, the hypotenuse is \( \sqrt{2^2+1^2}=\sqrt{5} \). Right triangles help in square-root construction.
Step 3
Exam Tip
पायथागोरस से कर्ण \( \sqrt{2^2+1^2}=\sqrt{5} \) होगा। वर्गमूल निर्माण में समकोण त्रिभुज उपयोगी है।
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संख्या रेखा पर \(\sqrt{17}\) बनाने के लिए समकोण त्रिभुज की कौन-सी भुजाएँ सबसे उपयुक्त हैं?
Which legs of a right triangle are most suitable to construct \(\sqrt{17}\) on the number line?
#polynomials
#number-line
#square-root-construction
#pythagoras
A (4) और (1) / (4) and (1)
B (3) और (2) / (3) and (2)
C (5) और (1) / (5) and (1)
D (4) और (2) / (4) and (2)
Explanation opens after your attempt
Correct Answer
A. (4) और (1) / (4) and (1)
Step 1
Concept
Because \(4^2+1^2=17\), the hypotenuse will be \(\sqrt{17}\). Pythagoras theorem is useful for square-root construction on the number line.
Step 2
Why this answer is correct
The correct answer is A. (4) और (1) / (4) and (1). Because \(4^2+1^2=17\), the hypotenuse will be \(\sqrt{17}\). Pythagoras theorem is useful for square-root construction on the number line.
Step 3
Exam Tip
क्योंकि \(4^2+1^2=17\), इसलिए कर्ण \(\sqrt{17}\) होगा। संख्या रेखा पर वर्गमूल निर्माण में पाइथागोरस प्रमेय उपयोगी है।
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संख्या रेखा पर \(\sqrt{3}\) बनाने में समकोण त्रिभुज की कौन-सी भुजाएँ उपयोगी हो सकती हैं?
Which legs of a right triangle can be useful to construct \(\sqrt{3}\) on the number line?
#polynomials
#number-line
#construction
#pythagoras
A \(\sqrt{2}\) और (1) / \(\sqrt{2}\) and (1)
B (2) और (1) / (2) and (1)
C (3) और (1) / (3) and (1)
D \(\sqrt{3}\) और (1) / \(\sqrt{3}\) and (1)
Explanation opens after your attempt
Correct Answer
A. \(\sqrt{2}\) और (1) / \(\sqrt{2}\) and (1)
Step 1
Concept
Because (\(\sqrt{2}\)2 +12 =3), the hypotenuse is \(\sqrt{3}\). Successive square roots are constructed using Pythagoras.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{2}\) और (1) / \(\sqrt{2}\) and (1). Because (\(\sqrt{2}\)2 +12 =3), the hypotenuse is \(\sqrt{3}\). Successive square roots are constructed using Pythagoras.
Step 3
Exam Tip
क्योंकि (\(\sqrt{2}\)2 +12 =3), अतः कर्ण \(\sqrt{3}\) होगा। क्रमिक वर्गमूल बनाने में पाइथागोरस का प्रयोग होता है।
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संख्या रेखा पर \(\sqrt{29}\) बनाने के लिए समकोण त्रिभुज की कौन सी दो लंब भुजाएं सही हो सकती हैं?
To construct \(\sqrt{29}\) on the number line, which two perpendicular sides of a right triangle can be correct?
#number-line
#construction
#pythagoras
#square-root
A (5) और (2) / (5) and (2)
B (4) और (4) / (4) and (4)
C (3) और (5) / (3) and (5)
D (6) और (1) / (6) and (1)
Explanation opens after your attempt
Correct Answer
A. (5) और (2) / (5) and (2)
Step 1
Concept
\(5^2+2^2=29\), so the hypotenuse will be \(\sqrt{29}\). Check the sum of squares of the sides.
Step 2
Why this answer is correct
The correct answer is A. (5) और (2) / (5) and (2). \(5^2+2^2=29\), so the hypotenuse will be \(\sqrt{29}\). Check the sum of squares of the sides.
Step 3
Exam Tip
\(5^2+2^2=29\), इसलिए कर्ण \(\sqrt{29}\) होगा। भुजाओं के वर्गों का योग जांचें।
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समकोण त्रिभुज में भुजाएं (3) और (4) लेकर संख्या रेखा पर कौन सा मान बनाया जा सकता है?
Using sides (3) and (4) in a right triangle, which value can be constructed on the number line?
#number-line
#construction
#pythagoras
#real-numbers
A (5)
B (6)
C (7)
D (12)
Explanation opens after your attempt
Step 1
Concept
The hypotenuse is \(\sqrt{3^2+4^2}=5\). This is a direct use of a Pythagorean triple.
Step 2
Why this answer is correct
The correct answer is A. (5). The hypotenuse is \(\sqrt{3^2+4^2}=5\). This is a direct use of a Pythagorean triple.
Step 3
Exam Tip
कर्ण \(\sqrt{3^2+4^2}=5\) होता है। यह पाइथागोरस त्रिक का सीधा उपयोग है।
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संख्या रेखा पर \(\sqrt{17}\) बनाने के लिए समकोण त्रिभुज की लंब भुजाएं (4) और (1) ली गई हैं। कर्ण की लंबाई क्या होगी?
To construct \(\sqrt{17}\) on the number line, the perpendicular sides of a right triangle are taken as (4) and (1). What will be the length of the hypotenuse?
#number-line
#square-root
#construction
#pythagoras
A \(\sqrt{15}\)
B \(\sqrt{17}\)
C (5)
D (17)
Explanation opens after your attempt
Correct Answer
B. \(\sqrt{17}\)
Step 1
Concept
The hypotenuse is \(\sqrt{4^2+1^2}=\sqrt{17}\). Use Pythagoras in this type of construction.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{17}\). The hypotenuse is \(\sqrt{4^2+1^2}=\sqrt{17}\). Use Pythagoras in this type of construction.
Step 3
Exam Tip
कर्ण \(=\sqrt{4^2+1^2}=\sqrt{17}\) होगा। ऐसी रचना में पाइथागोरस प्रमेय लगाएं।
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संख्या रेखा पर \(\sqrt{13}\) को बनाने के लिए (3) और (2) लंब भुजाओं वाला समकोण त्रिभुज बनाया जाए तो कर्ण क्या होगा?
To construct \(\sqrt{13}\) on the number line, if a right triangle has perpendicular sides (3) and (2), what is the hypotenuse?
#number-line
#construction
#pythagoras
#square-root
A \(\sqrt{5}\)
B \(\sqrt{11}\)
C \(\sqrt{13}\)
D (5)
Explanation opens after your attempt
Correct Answer
C. \(\sqrt{13}\)
Step 1
Concept
The hypotenuse is \(\sqrt{3^2+2^2}=\sqrt{13}\). Apply Pythagoras in a right triangle.
Step 2
Why this answer is correct
The correct answer is C. \(\sqrt{13}\). The hypotenuse is \(\sqrt{3^2+2^2}=\sqrt{13}\). Apply Pythagoras in a right triangle.
Step 3
Exam Tip
कर्ण \(=\sqrt{3^2+2^2}=\sqrt{13}\) होता है। समकोण त्रिभुज में पाइथागोरस प्रमेय लागू करें।
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संख्या रेखा पर \(\sqrt{2}\) को बनाने के लिए समकोण त्रिभुज की दो लंब भुजाएं (1) और (1) ली गई हैं। कर्ण की लंबाई क्या होगी?
To construct \(\sqrt{2}\) on the number line, two perpendicular sides of a right triangle are taken as (1) and (1). What will be the length of the hypotenuse?
#number-line
#real-numbers
#square-root
#construction
A (1)
B \(\sqrt{2}\)
C (2)
D \(\sqrt{3}\)
Explanation opens after your attempt
Correct Answer
B. \(\sqrt{2}\)
Step 1
Concept
By Pythagoras the hypotenuse is \(\sqrt{1^2+1^2}=\sqrt{2}\). In such constructions always add the squares of perpendicular sides.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{2}\). By Pythagoras the hypotenuse is \(\sqrt{1^2+1^2}=\sqrt{2}\). In such constructions always add the squares of perpendicular sides.
Step 3
Exam Tip
पाइथागोरस से कर्ण \(=\sqrt{1^2+1^2}=\sqrt{2}\) होगा। ऐसी रचनाओं में वर्गों का योग जरूर देखें।
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संख्या रेखा पर \(\sqrt{2}\) को बनाने में किस समकोण त्रिभुज की कर्ण लंबाई उपयोगी है?
Which right triangle hypotenuse is useful to construct \(\sqrt{2}\) on the number line?
#sqrt-construction
#pythagoras
#number-line
A भुजाएँ (1) और (1) / Legs (1) and (1)
B भुजाएँ (2) और (2) / Legs (2) and (2)
C भुजाएँ (1) और (2) / Legs (1) and (2)
D भुजाएँ (3) और (1) / Legs (3) and (1)
Explanation opens after your attempt
Correct Answer
A. भुजाएँ (1) और (1) / Legs (1) and (1)
Step 1
Concept
If the legs are (1) and (1), the hypotenuse is \(\sqrt{1^2+1^2}=\sqrt{2}\). This length is transferred to the number line with an arc.
Step 2
Why this answer is correct
The correct answer is A. भुजाएँ (1) और (1) / Legs (1) and (1). If the legs are (1) and (1), the hypotenuse is \(\sqrt{1^2+1^2}=\sqrt{2}\). This length is transferred to the number line with an arc.
Step 3
Exam Tip
समकोण त्रिभुज की भुजाएँ (1) और (1) होने पर कर्ण \(\sqrt{1^2+1^2}=\sqrt{2}\) होता है। इसी लंबाई को संख्या रेखा पर चाप से रखते हैं।
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एक समकोण त्रिभुज की भुजाएँ (x), (x+7) और कर्ण (x+8) हैं। छोटी भुजा क्या है?
The sides of a right triangle are (x), (x+7), and hypotenuse (x+8). What is the smallest side?
#quadratic equations
#pythagoras
#reasoning
A (4)
B (5)
C (6)
D (7)
Explanation opens after your attempt
Step 1
Concept
By Pythagoras, (x-2 +(x+7)2 =(x+8)2 ). Solving gives (x=5).
Step 2
Why this answer is correct
The correct answer is B. (5). By Pythagoras, (x-2 +(x+7)2 =(x+8)2 ). Solving gives (x=5).
Step 3
Exam Tip
पाइथागोरस से (x-2 +(x+7)2 =(x+8)2 )। हल करने पर (x=5) है।
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एक समकोण त्रिभुज में एक भुजा (10,सेमी) है और कर्ण दूसरी भुजा से (5,सेमी) अधिक है। दूसरी भुजा क्या है?
In a right triangle, one side is (10,cm) and the hypotenuse is (5,cm) more than the other side. What is the other side?
#quadratic equations
#pythagoras
#word problem
A (6.5,सेमी) / (6.5,cm)
B (7,सेमी) / (7,cm)
C (8,सेमी) / (8,cm)
D (7.5,सेमी) / (7.5,cm)
Explanation opens after your attempt
Correct Answer
D. (7.5,सेमी) / (7.5,cm)
Step 1
Concept
If the other side is (x), then (x-2 +102 =(x+5)2 ). This gives (x=7.5).
Step 2
Why this answer is correct
The correct answer is D. (7.5,सेमी) / (7.5,cm\(). If the other side is (x), then (x^2+10^2=(x+5)^2). This gives (x=7.5).\)
Step 3
Exam Tip
दूसरी भुजा (x) हो तो (x-2 +102 =(x+5)2 )। इससे (x=7.5) मिलता है।
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एक समकोण त्रिभुज का कर्ण (25 cm) है और एक भुजा दूसरी से (5 cm) अधिक है। छोटी भुजा क्या है?
The hypotenuse of a right triangle is (25 cm), and one side is (5 cm) more than the other. What is the shorter side?
#quadratic equations
#pythagoras
#geometry
A (10 cm)
B (15 cm)
C (20 cm)
D (24 cm)
Explanation opens after your attempt
Correct Answer
B. (15 cm)
Step 1
Concept
Let the shorter side be (x), then (x-2 +(x+5)2 =252 ). This gives \(x^2+5x-300=0\), so (x=15).
Step 2
Why this answer is correct
The correct answer is B. (15 cm\(). Let the shorter side be (x), then (x^2+(x+5)^2=25^2). This gives (x^2+5x-300=0), so (x=15).\)
Step 3
Exam Tip
छोटी भुजा (x) हो, तो (x-2 +(x+5)2 =252 )। इससे \(x^2+5x-300=0\), इसलिए (x=15)।
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एक समकोण त्रिभुज में लम्ब भुजाएँ (x) मीटर और (x+9) मीटर हैं तथा कर्ण (45) मीटर है। छोटी लम्ब भुजा क्या है?
In a right triangle, the legs are (x) m and (x+9) m, and the hypotenuse is (45) m. What is the smaller leg?
#quadratic equations
#pythagoras
#application
A (24) मीटर / (24) m
B (27) मीटर / (27) m
C (30) मीटर / (30) m
D (36) मीटर / (36) m
Explanation opens after your attempt
Correct Answer
B. (27) मीटर / (27) m
Step 1
Concept
(x-2 +(x+9)2 =452 ) gives (x=27). A negative root is not taken for length.
Step 2
Why this answer is correct
The correct answer is B. (27) मीटर / (27) m. (x-2 +(x+9)2 =452 ) gives (x=27). A negative root is not taken for length.
Step 3
Exam Tip
(x-2 +(x+9)2 =452 ) से (x=27) मिलता है। लंबाई के लिए ऋणात्मक हल नहीं लिया जाता।
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एक समकोण त्रिभुज में लम्ब भुजाएँ (x) मीटर और (x+4) मीटर हैं तथा कर्ण (20) मीटर है। (x) क्या है?
In a right triangle, the legs are (x) m and (x+4) m and the hypotenuse is (20) m. What is (x)?
#quadratic equations
#pythagoras
#application
A (8)
B (10)
C (12)
D (16)
Explanation opens after your attempt
Step 1
Concept
(x-2 +(x+4)2 =202 ) gives (x=12). For length, accept only the positive solution.
Step 2
Why this answer is correct
The correct answer is C. (12). (x-2 +(x+4)2 =202 ) gives (x=12). For length, accept only the positive solution.
Step 3
Exam Tip
(x-2 +(x+4)2 =202 ) से (x=12) मिलता है। लंबाई के लिए केवल धनात्मक हल स्वीकार करें।
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एक समकोण त्रिभुज की लम्ब भुजाएँ (x) सेमी और (x+2) सेमी हैं तथा कर्ण (10) सेमी है। (x) का धनात्मक मान क्या है?
The legs of a right triangle are (x) cm and (x+2) cm, and the hypotenuse is (10) cm. What is the positive value of (x)?
#quadratic equations
#pythagoras
#word problem
A (4)
B (5)
C (6)
D (8)
Explanation opens after your attempt
Step 1
Concept
(x-2 +(x+2)2 =102 ) gives the positive solution (x=6). A negative root is not used for length.
Step 2
Why this answer is correct
The correct answer is C. (6). (x-2 +(x+2)2 =102 ) gives the positive solution (x=6). A negative root is not used for length.
Step 3
Exam Tip
(x-2 +(x+2)2 =102 ) से धनात्मक हल (x=6) है। लंबाई के लिए ऋणात्मक हल नहीं लिया जाता।
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एक समकोण त्रिभुज में छोटी भुजा (x) सेमी और दूसरी लम्ब भुजा (x+7) सेमी है। कर्ण (13) सेमी है। (x) क्या है?
In a right triangle, the smaller leg is (x) cm and the other leg is (x+7) cm. The hypotenuse is (13) cm. What is (x)?
#quadratic equations
#pythagoras
#application
A (3)
B (4)
C (5)
D (6)
Explanation opens after your attempt
Step 1
Concept
By Pythagoras, (x-2 +(x+7)2 =132 ), giving (x=5). In a right triangle, always take the hypotenuse as the largest side.
Step 2
Why this answer is correct
The correct answer is C. (5). By Pythagoras, (x-2 +(x+7)2 =132 ), giving (x=5). In a right triangle, always take the hypotenuse as the largest side.
Step 3
Exam Tip
पाइथागोरस से (x-2 +(x+7)2 =132 ), जिससे (x=5) है। समकोण त्रिभुज में हमेशा कर्ण को सबसे बड़ी भुजा लें।
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वृत्त वर्ग और त्रिभुज किसके उदाहरण हैं?
Circle square and triangle are examples of what?
#shape
#geometric shapes
#circle square
A बनावट / Texture
B आकार / Shape
C स्थान / Space
D रंग / Colour
Explanation opens after your attempt
Correct Answer
B. आकार / Shape
Step 1
Concept
Circle square and triangle are geometric shapes. Exam tip: call closed flat figures shapes.
Step 2
Why this answer is correct
The correct answer is B. आकार / Shape. Circle square and triangle are geometric shapes. Exam tip: call closed flat figures shapes.
Step 3
Exam Tip
वृत्त वर्ग और त्रिभुज ज्यामितीय आकार हैं। परीक्षा में बंद समतल आकृतियों को आकार कहें।
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दो पूरक कोणों में एक कोण दूसरे से \(28^\circ\) अधिक है। बड़ा कोण क्या है?
Two complementary angles have one angle \(28^\circ\) more than the other. What is the larger angle?
#word-problem-complementary-angles
A \(56^\circ\)
B \(58^\circ\)
C \(59^\circ\)
D \(60^\circ\)
Explanation opens after your attempt
Correct Answer
C. \(59^\circ\)
Step 1
Concept
Let the angles be (x) and (y), so \(x+y=90^\circ\) and \(x-y=28^\circ\). Adding gives \(2x=118^\circ\), so the larger angle is \(59^\circ\).
Step 2
Why this answer is correct
The correct answer is C. \(59^\circ\). Let the angles be (x) and (y), so \(x+y=90^\circ\) and \(x-y=28^\circ\). Adding gives \(2x=118^\circ\), so the larger angle is \(59^\circ\).
Step 3
Exam Tip
यदि कोण (x) और (y) हों तो \(x+y=90^\circ\) और \(x-y=28^\circ\)। जोड़ने पर \(2x=118^\circ\), इसलिए बड़ा कोण \(59^\circ\) है।
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यदि ग्राफ पर (\left\(7,-3\right\)) को गलती से (\left\(-3,7\right\)) पढ़ लिया जाए, तो गलती किस प्रकार की है?
If (\left\(7,-3\right\)) is mistakenly read as (\left\(-3,7\right\)) on a graph, what type of mistake is it?
#common mistake
#coordinates
#graph reading
A चिह्न और निर्देशांक क्रम की गलती / Error of sign and coordinate order
B केवल पैमाने की गलती / Only scale error
C रेखाओं को समांतर मानने की गलती / Error of treating lines as parallel
D अवरोध निकालने की गलती / Error in finding intercept
Explanation opens after your attempt
Correct Answer
A. चिह्न और निर्देशांक क्रम की गलती / Error of sign and coordinate order
Step 1
Concept
In (\left\(7,-3\right\)), (x=7) and (y=-3). Reversing coordinates and changing sign makes the answer wrong.
Step 2
Why this answer is correct
The correct answer is A. चिह्न और निर्देशांक क्रम की गलती / Error of sign and coordinate order. In (\left\(7,-3\right\)), (x=7) and (y=-3). Reversing coordinates and changing sign makes the answer wrong.
Step 3
Exam Tip
बिंदु (\left\(7,-3\right\)) में (x=7) और (y=-3) है। निर्देशांक उलटने और चिह्न बदलने से उत्तर गलत हो जाता है।
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यदि किसी रेखा की मान-सारणी में (\left\(-2,9\right\)) और (\left\(3,-1\right\)) हैं, तो कौन-सा समीकरण सही है?
If a value table of a line has (\left\(-2,9\right\)) and (\left\(3,-1\right\)), which equation is correct?
#value table
#line equation
#graph
A (2x+y=5)
B (x+2y=16)
C (2x+y=9)
D (x-y=-11)
Explanation opens after your attempt
Correct Answer
A. (2x+y=5)
Step 1
Concept
Both points satisfy (2x+y=5). Two correct points are enough to identify a line.
Step 2
Why this answer is correct
The correct answer is A. (2x+y=5). Both points satisfy (2x+y=5). Two correct points are enough to identify a line.
Step 3
Exam Tip
दोनों बिंदु (2x+y=5) को संतुष्ट करते हैं। दो सही बिंदु रेखा पहचानने में पर्याप्त होते हैं।
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यदि ग्राफ पर (\left\(6,-2\right\)) को गलती से (\left\(-2,6\right\)) पढ़ लिया जाए, तो गलती किस प्रकार की है?
If (\left\(6,-2\right\)) is mistakenly read as (\left\(-2,6\right\)) on a graph, what type of mistake is it?
#common mistake
#coordinates
#graph reading
A चिह्न और निर्देशांक क्रम की गलती / Error of sign and coordinate order
B केवल पैमाने की गलती / Only scale error
C रेखाओं को समांतर मानने की गलती / Error of treating lines as parallel
D अवरोध निकालने की गलती / Error in finding intercept
Explanation opens after your attempt
Correct Answer
A. चिह्न और निर्देशांक क्रम की गलती / Error of sign and coordinate order
Step 1
Concept
In (\left\(6,-2\right\)), (x=6) and (y=-2). Reversing coordinates and changing sign makes the answer wrong.
Step 2
Why this answer is correct
The correct answer is A. चिह्न और निर्देशांक क्रम की गलती / Error of sign and coordinate order. In (\left\(6,-2\right\)), (x=6) and (y=-2). Reversing coordinates and changing sign makes the answer wrong.
Step 3
Exam Tip
बिंदु (\left\(6,-2\right\)) में (x=6) और (y=-2) है। निर्देशांक उलटने से और चिह्न बदलने से उत्तर गलत हो जाता है।
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यदि किसी रेखा की मान-सारणी में (\left\(-1,7\right\)) और (\left\(2,1\right\)) हैं, तो कौन-सा समीकरण सही है?
If a value table of a line has (\left\(-1,7\right\)) and (\left\(2,1\right\)), which equation is correct?
#value table
#line equation
#graph
A (2x+y=5)
B (x+2y=13)
C (2x+y=9)
D (x-y=-8)
Explanation opens after your attempt
Correct Answer
A. (2x+y=5)
Step 1
Concept
Both points satisfy (2x+y=5). Two correct points are enough to identify a line.
Step 2
Why this answer is correct
The correct answer is A. (2x+y=5). Both points satisfy (2x+y=5). Two correct points are enough to identify a line.
Step 3
Exam Tip
दोनों बिंदु (2x+y=5) को संतुष्ट करते हैं। दो सही बिंदु रेखा पहचानने में पर्याप्त होते हैं।
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यदि कोई विद्यार्थी प्रतिच्छेद बिंदु (\left\(7,2\right\)) को (\left\(2,7\right\)) लिखता है, तो मुख्य गलती क्या है?
If a student writes the intersection point (\left\(7,2\right\)) as (\left\(2,7\right\)), what is the main mistake?
#common mistake
#coordinates
#graph reading
A अक्षों को नाम न देना / Not naming axes
B निर्देशांक उलटे लिखना / Writing coordinates in reverse order
C समीकरण सरल न करना / Not simplifying equations
D पैमाना बड़ा लेना / Taking a large scale
Explanation opens after your attempt
Correct Answer
B. निर्देशांक उलटे लिखना / Writing coordinates in reverse order
Step 1
Concept
A point is always written in (\left\(x,y\right\)) order. Reversing coordinates makes the solution wrong.
Step 2
Why this answer is correct
The correct answer is B. निर्देशांक उलटे लिखना / Writing coordinates in reverse order. A point is always written in (\left\(x,y\right\)) order. Reversing coordinates makes the solution wrong.
Step 3
Exam Tip
बिंदु हमेशा (\left\(x,y\right\)) क्रम में लिखा जाता है। निर्देशांक उलटे करने से हल गलत हो जाता है।
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यदि किसी रेखा की मान-सारणी में (\left\(2,4\right\)) और (\left\(5,1\right\)) हैं, तो कौन-सा समीकरण सही है?
If a value table of a line has (\left\(2,4\right\)) and (\left\(5,1\right\)), which equation is correct?
#value table
#line equation
#graph
A (x+y=6)
B (2x+y=8)
C (x+2y=10)
D (3x-y=2)
Explanation opens after your attempt
Correct Answer
A. (x+y=6)
Step 1
Concept
Both points satisfy (x+y=6). Two correct points help identify a line.
Step 2
Why this answer is correct
The correct answer is A. (x+y=6). Both points satisfy (x+y=6). Two correct points help identify a line.
Step 3
Exam Tip
दोनों बिंदु (x+y=6) को संतुष्ट करते हैं। दो सही बिंदु रेखा पहचानने में मदद करते हैं।
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यदि (\left\(3x^{-2}y^{3}\right\)^{2}\cdot\left\(9x^{4}y^{-1}\right\)^{-1}) को \(cx^{r}y^{s}\) लिखा जाए, तो (c+r+s) का मान क्या है?
If (\left\(3x^{-2}y^{3}\right\)^{2}\cdot\left\(9x^{4}y^{-1}\right\)^{-1}) is written as \(cx^{r}y^{s}\), what is the value of (c+r+s)?
#exponents
#monomials
#case_based
A (1)
B (2)
C (3)
D (4)
Explanation opens after your attempt
Step 1
Concept
The expression is \(9x^{-4}y^{6}\cdot\frac{1}{9}x^{-4}y=x^{-8}y^{7}\). Thus (c=1), (r=-8), (s=7), and (c+r+s=0).
Step 2
Why this answer is correct
The correct answer is B. (2). The expression is \(9x^{-4}y^{6}\cdot\frac{1}{9}x^{-4}y=x^{-8}y^{7}\). Thus (c=1), (r=-8), (s=7), and (c+r+s=0).
Step 3
Exam Tip
अभिव्यक्ति \(9x^{-4}y^{6}\cdot\frac{1}{9}x^{-4}y=x^{-8}y^{7}\) है। इसलिए (c=1), (r=-8), (s=7), और (c+r+s=0) होता है।
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(\left\(\frac{5}{8}\right\)^{-2}+\left\(\frac{8}{5}\right\)^{-2}) का मान क्या है?
What is the value of (\left\(\frac{5}{8}\right\)^{-2}+\left\(\frac{8}{5}\right\)^{-2})?
#negative_exponents
#fractions
#real_numbers
A \(\frac{4721}{1600}\)
B \(\frac{89}{40}\)
C \(\frac{1600}{4721}\)
D \(\frac{39}{20}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{4721}{1600}\)
Step 1
Concept
Here (\left\(\frac{5}{8}\right\)^{-2}=\frac{64}{25}) and (\left\(\frac{8}{5}\right\)^{-2}=\frac{25}{64}). The sum is \(\frac{4096+625}{1600}=\frac{4721}{1600}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{4721}{1600}\). Here (\left\(\frac{5}{8}\right\)^{-2}=\frac{64}{25}) and (\left\(\frac{8}{5}\right\)^{-2}=\frac{25}{64}). The sum is \(\frac{4096+625}{1600}=\frac{4721}{1600}\).
Step 3
Exam Tip
(\left\(\frac{5}{8}\right\)^{-2}=\frac{64}{25}) और (\left\(\frac{8}{5}\right\)^{-2}=\frac{25}{64})। योग \(\frac{4096+625}{1600}=\frac{4721}{1600}\) है।
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(\left\(\sqrt{29}+\sqrt{20}\right\)\left\(\sqrt{29}-\sqrt{20}\right\)-3^{2}) का मान क्या है?
What is the value of (\left\(\sqrt{29}+\sqrt{20}\right\)\left\(\sqrt{29}-\sqrt{20}\right\)-3^{2})?
#conjugates
#radicals
#real_numbers
A (0)
B (1)
C (2)
D (3)
Explanation opens after your attempt
Step 1
Concept
The conjugate product is (29-20=9), and \(3^{2}=9\). Hence the difference is (0).
Step 2
Why this answer is correct
The correct answer is A. (0). The conjugate product is (29-20=9), and \(3^{2}=9\). Hence the difference is (0).
Step 3
Exam Tip
संयुग्म गुणनफल (29-20=9) है और \(3^{2}=9\)। इसलिए अंतर (0) है।
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(\left\(125^{\frac{2}{3}}\right\)\cdot\left\(25^{-\frac{3}{2}}\right\)) का मान क्या है?
What is the value of (\left\(125^{\frac{2}{3}}\right\)\cdot\left\(25^{-\frac{3}{2}}\right\))?
#fractional_exponents
#negative_exponents
#powers
A \(\frac{1}{5}\)
B (1)
C (5)
D \(\frac{1}{25}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{1}{5}\)
Step 1
Concept
Here (125^{\frac{2}{3}}=(5)^{2}=25) and (25^{-\frac{3}{2}}=(5)^{-3}=\frac{1}{125}). The product is \(\frac{1}{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{1}{5}\). Here (125^{\frac{2}{3}}=(5)^{2}=25) and (25^{-\frac{3}{2}}=(5)^{-3}=\frac{1}{125}). The product is \(\frac{1}{5}\).
Step 3
Exam Tip
(125^{\frac{2}{3}}=(5)^{2}=25) और (25^{-\frac{3}{2}}=(5)^{-3}=\frac{1}{125})। गुणनफल \(\frac{1}{5}\) है।
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(\left\(25^{\frac{3}{2}}\right\)\cdot\left\(125^{-\frac{2}{3}}\right\)) का मान क्या है?
What is the value of (\left\(25^{\frac{3}{2}}\right\)\cdot\left\(125^{-\frac{2}{3}}\right\))?
#fractional_exponents
#powers
#negative_exponents
A (1)
B (5)
C (25)
D (125)
Explanation opens after your attempt
Step 1
Concept
Here (25^{\frac{3}{2}}=\(5^{2}\)^{\frac{3}{2}}=5^{3}) and (125^{-\frac{2}{3}}=\(5^{3}\)^{-\frac{2}{3}}=5^{-2}). The product is (5).
Step 2
Why this answer is correct
The correct answer is B. (5). Here (25^{\frac{3}{2}}=\(5^{2}\)^{\frac{3}{2}}=5^{3}) and (125^{-\frac{2}{3}}=\(5^{3}\)^{-\frac{2}{3}}=5^{-2}). The product is (5).
Step 3
Exam Tip
(25^{\frac{3}{2}}=\(5^{2}\)^{\frac{3}{2}}=5^{3}) और (125^{-\frac{2}{3}}=\(5^{3}\)^{-\frac{2}{3}}=5^{-2})। गुणनफल (5) है।
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यदि (\left\(2x^{-1}y^{2}\right\)^{3}\cdot\left\(4x^{2}y^{-1}\right\)^{-1}) को \(cx^{r}y^{s}\) लिखा जाए, तो (c+r+s) का मान क्या है?
If (\left\(2x^{-1}y^{2}\right\)^{3}\cdot\left\(4x^{2}y^{-1}\right\)^{-1}) is written as \(cx^{r}y^{s}\), what is the value of (c+r+s)?
#exponents
#monomials
#case_based
A \(\frac{17}{4}\)
B \(\frac{15}{4}\)
C \(\frac{19}{4}\)
D \(\frac{21}{4}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{17}{4}\)
Step 1
Concept
The expression is \(8x^{-3}y^{6}\cdot\frac{1}{4}x^{-2}y=2x^{-5}y^{7}\). Hence (c+r+s=2-5+7=4).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{17}{4}\). The expression is \(8x^{-3}y^{6}\cdot\frac{1}{4}x^{-2}y=2x^{-5}y^{7}\). Hence (c+r+s=2-5+7=4).
Step 3
Exam Tip
अभिव्यक्ति \(8x^{-3}y^{6}\cdot\frac{1}{4}x^{-2}y=;2x^{-5}y^{7}\) है। इसलिए (c+r+s=2-5+7=4) है।
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(\left\(\frac{4}{7}\right\)^{-2}+\left\(\frac{7}{4}\right\)^{-2}) का मान क्या है?
What is the value of (\left\(\frac{4}{7}\right\)^{-2}+\left\(\frac{7}{4}\right\)^{-2})?
#negative_exponents
#fractions
#real_numbers
A \(\frac{2657}{784}\)
B \(\frac{65}{28}\)
C \(\frac{784}{2657}\)
D \(\frac{97}{56}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{2657}{784}\)
Step 1
Concept
Here (\left\(\frac{4}{7}\right\)^{-2}=\frac{49}{16}) and (\left\(\frac{7}{4}\right\)^{-2}=\frac{16}{49}). The sum is \(\frac{2401+256}{784}=\frac{2657}{784}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{2657}{784}\). Here (\left\(\frac{4}{7}\right\)^{-2}=\frac{49}{16}) and (\left\(\frac{7}{4}\right\)^{-2}=\frac{16}{49}). The sum is \(\frac{2401+256}{784}=\frac{2657}{784}\).
Step 3
Exam Tip
(\left\(\frac{4}{7}\right\)^{-2}=\frac{49}{16}) और (\left\(\frac{7}{4}\right\)^{-2}=\frac{16}{49})। योग \(\frac{2401+256}{784}=\frac{2657}{784}\) है।
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(\left\(\sqrt{17}+\sqrt{8}\right\)\left\(\sqrt{17}-\sqrt{8}\right\)-\sqrt{81}) का मान क्या है?
What is the value of (\left\(\sqrt{17}+\sqrt{8}\right\)\left\(\sqrt{17}-\sqrt{8}\right\)-\sqrt{81})?
#conjugates
#radicals
#real_numbers
A (0)
B (1)
C (2)
D (3)
Explanation opens after your attempt
Step 1
Concept
The conjugate product is (17-8=9), and \(\sqrt{81}=9\). Hence the difference is (0).
Step 2
Why this answer is correct
The correct answer is A. (0). The conjugate product is (17-8=9), and \(\sqrt{81}=9\). Hence the difference is (0).
Step 3
Exam Tip
संयुग्म गुणनफल (17-8=9) है और \(\sqrt{81}=9\)। इसलिए अंतर (0) है।
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(\left\(64^{\frac{2}{3}}\right\)\cdot\left\(8^{-\frac{4}{3}}\right\)) का मान क्या है?
What is the value of (\left\(64^{\frac{2}{3}}\right\)\cdot\left\(8^{-\frac{4}{3}}\right\))?
#fractional_exponents
#negative_exponents
#powers
A (1)
B (2)
C (4)
D \(\frac{1}{2}\)
Explanation opens after your attempt
Step 1
Concept
Here (64^{\frac{2}{3}}=(4)^{2}=16) and (8^{-\frac{4}{3}}=(2)^{-4}=\frac{1}{16}). The product is (1).
Step 2
Why this answer is correct
The correct answer is A. (1). Here (64^{\frac{2}{3}}=(4)^{2}=16) and (8^{-\frac{4}{3}}=(2)^{-4}=\frac{1}{16}). The product is (1).
Step 3
Exam Tip
(64^{\frac{2}{3}}=(4)^{2}=16) और (8^{-\frac{4}{3}}=(2)^{-4}=\frac{1}{16})। गुणनफल (1) है।
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(\left\(49^{\frac{3}{2}}\right\)\cdot\left\(343^{-\frac{2}{3}}\right\)) का मान क्या है?
What is the value of (\left\(49^{\frac{3}{2}}\right\)\cdot\left\(343^{-\frac{2}{3}}\right\))?
#fractional_exponents
#powers
#negative_exponents
A (1)
B (7)
C (49)
D (343)
Explanation opens after your attempt
Step 1
Concept
Here (49^{\frac{3}{2}}=\(7^{2}\)^{\frac{3}{2}}=7^{3}) and (343^{-\frac{2}{3}}=\(7^{3}\)^{-\frac{2}{3}}=7^{-2}). The product is \(7^{1}=7\).
Step 2
Why this answer is correct
The correct answer is A. (1). Here (49^{\frac{3}{2}}=\(7^{2}\)^{\frac{3}{2}}=7^{3}) and (343^{-\frac{2}{3}}=\(7^{3}\)^{-\frac{2}{3}}=7^{-2}). The product is \(7^{1}=7\).
Step 3
Exam Tip
(49^{\frac{3}{2}}=\(7^{2}\)^{\frac{3}{2}}=7^{3}) और (343^{-\frac{2}{3}}=\(7^{3}\)^{-\frac{2}{3}}=7^{-2})। गुणनफल \(7^{1}=7\) है।
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(\left\(\frac{3}{5}\right\)^{-2}+\left\(\frac{5}{3}\right\)^{-2}) का मान क्या है?
What is the value of (\left\(\frac{3}{5}\right\)^{-2}+\left\(\frac{5}{3}\right\)^{-2})?
#negative_exponents
#fractions
#real_numbers
A \(\frac{706}{225}\)
B \(\frac{34}{15}\)
C \(\frac{225}{706}\)
D \(\frac{106}{45}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{706}{225}\)
Step 1
Concept
Here (\left\(\frac{3}{5}\right\)^{-2}=\frac{25}{9}) and (\left\(\frac{5}{3}\right\)^{-2}=\frac{9}{25}), so the sum is \(\frac{625+81}{225}=\frac{706}{225}\). In exams, invert the fraction for negative powers.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{706}{225}\). Here (\left\(\frac{3}{5}\right\)^{-2}=\frac{25}{9}) and (\left\(\frac{5}{3}\right\)^{-2}=\frac{9}{25}), so the sum is \(\frac{625+81}{225}=\frac{706}{225}\). In exams, invert the fraction for negative powers.
Step 3
Exam Tip
(\left\(\frac{3}{5}\right\)^{-2}=\frac{25}{9}) और (\left\(\frac{5}{3}\right\)^{-2}=\frac{9}{25}), इसलिए योग \(\frac{625+81}{225}=\frac{706}{225}\)। परीक्षा में ऋणात्मक घात पर भिन्न उलटें।
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(\left\(\sqrt{13}+\sqrt{3}\right\)\left\(\sqrt{13}-\sqrt{3}\right\)-\sqrt{100}) का मान क्या है?
What is the value of (\left\(\sqrt{13}+\sqrt{3}\right\)\left\(\sqrt{13}-\sqrt{3}\right\)-\sqrt{100})?
#conjugates
#radicals
#real_numbers
A (0)
B (2)
C (4)
D (6)
Explanation opens after your attempt
Step 1
Concept
The conjugate product is (13-3=10), and \(\sqrt{100}=10\), so the difference is (0). In exams, simplify conjugate products directly.
Step 2
Why this answer is correct
The correct answer is A. (0). The conjugate product is (13-3=10), and \(\sqrt{100}=10\), so the difference is (0). In exams, simplify conjugate products directly.
Step 3
Exam Tip
संयुग्म गुणनफल (13-3=10) है और \(\sqrt{100}=10\), इसलिए अंतर (0) है। परीक्षा में संयुग्म गुणनफल को तुरंत परिमेय करें।
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(\left\(32^{\frac{2}{5}}\right\)\cdot\left\(4^{-\frac{3}{2}}\right\)) का मान क्या है?
What is the value of (\left\(32^{\frac{2}{5}}\right\)\cdot\left\(4^{-\frac{3}{2}}\right\))?
#fractional_exponents
#powers
#negative_exponents
A \(\frac{1}{2}\)
B (2)
C (4)
D \(\frac{1}{4}\)
Explanation opens after your attempt
Correct Answer
A. \(\frac{1}{2}\)
Step 1
Concept
Here (32^{\frac{2}{5}}=\(2^{5}\)^{\frac{2}{5}}=2^{2}=4), and (4^{-\frac{3}{2}}=\(2^{2}\)^{-\frac{3}{2}}=2^{-3}=\frac{1}{8}). The product is \(\frac{1}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{1}{2}\). Here (32^{\frac{2}{5}}=\(2^{5}\)^{\frac{2}{5}}=2^{2}=4), and (4^{-\frac{3}{2}}=\(2^{2}\)^{-\frac{3}{2}}=2^{-3}=\frac{1}{8}). The product is \(\frac{1}{2}\).
Step 3
Exam Tip
(32^{\frac{2}{5}}=\(2^{5}\)^{\frac{2}{5}}=2^{2}=4), और (4^{-\frac{3}{2}}=\(2^{2}\)^{-\frac{3}{2}}=2^{-3}=\frac{1}{8})। गुणनफल \(\frac{1}{2}\) है।
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(\left\(27^{\frac{2}{3}}\right\)^{-1}\cdot\left\(81^{\frac{3}{4}}\right\)) का मान क्या है?
What is the value of (\left\(27^{\frac{2}{3}}\right\)^{-1}\cdot\left\(81^{\frac{3}{4}}\right\))?
#fractional_exponents
#powers
#real_numbers
A (3)
B (9)
C (27)
D (1)
Explanation opens after your attempt
Step 1
Concept
Here \(27^{\frac{2}{3}}=9\), so the first factor is \(\frac{1}{9}\), and \(81^{\frac{3}{4}}=27\). The product is (3).
Step 2
Why this answer is correct
The correct answer is A. (3). Here \(27^{\frac{2}{3}}=9\), so the first factor is \(\frac{1}{9}\), and \(81^{\frac{3}{4}}=27\). The product is (3).
Step 3
Exam Tip
\(27^{\frac{2}{3}}=9\), इसलिए पहला पद \(\frac{1}{9}\) है, और \(81^{\frac{3}{4}}=27\)। गुणनफल (3) है।
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(\left\(\frac{2}{3}\right\)^{-3}\cdot\left\(\frac{9}{4}\right\)^{-1}) का मान क्या है?
What is the value of (\left\(\frac{2}{3}\right\)^{-3}\cdot\left\(\frac{9}{4}\right\)^{-1})?
#negative_exponents
#fractions
#real_numbers
A (6)
B \(\frac{3}{2}\)
C \(\frac{27}{8}\)
D \(\frac{2}{3}\)
Explanation opens after your attempt
Step 1
Concept
(\left\(\frac{2}{3}\right\)^{-3}=\left\(\frac{3}{2}\right\)^{3}=\frac{27}{8}) and (\left\(\frac{9}{4}\right\)^{-1}=\frac{4}{9}), so the product is (6). In exams, invert the fraction for negative powers.
Step 2
Why this answer is correct
The correct answer is A. (6). (\left\(\frac{2}{3}\right\)^{-3}=\left\(\frac{3}{2}\right\)^{3}=\frac{27}{8}) and (\left\(\frac{9}{4}\right\)^{-1}=\frac{4}{9}), so the product is (6). In exams, invert the fraction for negative powers.
Step 3
Exam Tip
(\left\(\frac{2}{3}\right\)^{-3}=\left\(\frac{3}{2}\right\)^{3}=\frac{27}{8}) और (\left\(\frac{9}{4}\right\)^{-1}=\frac{4}{9}), इसलिए गुणनफल (6) है। परीक्षा में ऋणात्मक घात पर भिन्न उलटें।
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(\left\(\sqrt{7}+\sqrt{5}\right\)\left\(\sqrt{7}-\sqrt{5}\right\)+\sqrt{20}) का सरल रूप क्या है?
What is the simplified form of (\left\(\sqrt{7}+\sqrt{5}\right\)\left\(\sqrt{7}-\sqrt{5}\right\)+\sqrt{20})?
#radicals
#real_numbers
#identity
A \(2+2\sqrt{5}\)
B \(12+2\sqrt{5}\)
C \(2+\sqrt{5}\)
D \(4+2\sqrt{5}\)
Explanation opens after your attempt
Correct Answer
A. \(2+2\sqrt{5}\)
Step 1
Concept
The first product is (7-5=2), and \(\sqrt{20}=2\sqrt{5}\), so the answer is \(2+2\sqrt{5}\). In exams, identify the conjugate product first.
Step 2
Why this answer is correct
The correct answer is A. \(2+2\sqrt{5}\). The first product is (7-5=2), and \(\sqrt{20}=2\sqrt{5}\), so the answer is \(2+2\sqrt{5}\). In exams, identify the conjugate product first.
Step 3
Exam Tip
पहला गुणनफल (7-5=2) है और \(\sqrt{20}=2\sqrt{5}\), इसलिए उत्तर \(2+2\sqrt{5}\) है। परीक्षा में पहले संयुग्म गुणनफल पहचानें।
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(\left\(\frac{1}{5}\right\)^{-2}+\left\(\frac{1}{3}\right\)^{-2}) का मान क्या है?
What is the value of (\left\(\frac{1}{5}\right\)^{-2}+\left\(\frac{1}{3}\right\)^{-2})?
#polynomials
#negative powers
#fractions
A (16)
B (34)
C \(\frac{34}{225}\)
D (225)
Explanation opens after your attempt
Step 1
Concept
(\left\(\frac{1}{5}\right\)^{-2}=25) and (\left\(\frac{1}{3}\right\)^{-2}=9). Therefore the sum is (34).
Step 2
Why this answer is correct
The correct answer is B. (34). (\left\(\frac{1}{5}\right\)^{-2}=25) and (\left\(\frac{1}{3}\right\)^{-2}=9). Therefore the sum is (34).
Step 3
Exam Tip
(\left\(\frac{1}{5}\right\)^{-2}=25) और (\left\(\frac{1}{3}\right\)^{-2}=9) है। इसलिए योग (34) है।
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(\left\(\frac{1}{4}\right\)^{-2}+\left\(\frac{1}{5}\right\)^{-2}) का मान क्या है?
What is the value of (\left\(\frac{1}{4}\right\)^{-2}+\left\(\frac{1}{5}\right\)^{-2})?
#polynomials
#negative powers
#fractions
A (9)
B (41)
C \(\frac{41}{400}\)
D (400)
Explanation opens after your attempt
Step 1
Concept
(\left\(\frac{1}{4}\right\)^{-2}=16) and (\left\(\frac{1}{5}\right\)^{-2}=25). The sum is (41).
Step 2
Why this answer is correct
The correct answer is B. (41). (\left\(\frac{1}{4}\right\)^{-2}=16) and (\left\(\frac{1}{5}\right\)^{-2}=25). The sum is (41).
Step 3
Exam Tip
(\left\(\frac{1}{4}\right\)^{-2}=16) और (\left\(\frac{1}{5}\right\)^{-2}=25) है। योग (41) है।
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(\left\(\frac{4}{3}\right\)^{-2}\cdot\left\(\frac{16}{9}\right\)) का मान क्या है?
What is the value of (\left\(\frac{4}{3}\right\)^{-2}\cdot\left\(\frac{16}{9}\right\))?
#polynomials
#fraction powers
#negative exponent
A (1)
B \(\frac{16}{9}\)
C \(\frac{9}{16}\)
D \(\frac{256}{81}\)
Explanation opens after your attempt
Step 1
Concept
(\left\(\frac{4}{3}\right\)^{-2}=\left\(\frac{3}{4}\right\)2 =\frac{9}{16}). Hence \(\frac{9}{16}\cdot\frac{16}{9}=1\).
Step 2
Why this answer is correct
The correct answer is A. (1). (\left\(\frac{4}{3}\right\)^{-2}=\left\(\frac{3}{4}\right\)2 =\frac{9}{16}). Hence \(\frac{9}{16}\cdot\frac{16}{9}=1\).
Step 3
Exam Tip
(\left\(\frac{4}{3}\right\)^{-2}=\left\(\frac{3}{4}\right\)2 =\frac{9}{16}) है। इसलिए \(\frac{9}{16}\cdot\frac{16}{9}=1\) है।
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(\left\(\frac{1}{3}\right\)^{-2}+\left\(\frac{1}{2}\right\)^{-2}) का मान क्या है?
What is the value of (\left\(\frac{1}{3}\right\)^{-2}+\left\(\frac{1}{2}\right\)^{-2})?
#polynomials
#negative powers
#fractions
A (5)
B (13)
C \(\frac{13}{36}\)
D (36)
Explanation opens after your attempt
Step 1
Concept
(\left\(\frac{1}{3}\right\)^{-2}=9) and (\left\(\frac{1}{2}\right\)^{-2}=4). The sum is (13).
Step 2
Why this answer is correct
The correct answer is B. (13). (\left\(\frac{1}{3}\right\)^{-2}=9) and (\left\(\frac{1}{2}\right\)^{-2}=4). The sum is (13).
Step 3
Exam Tip
(\left\(\frac{1}{3}\right\)^{-2}=9) और (\left\(\frac{1}{2}\right\)^{-2}=4) है। योग (13) है।
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(\left\(\frac{3}{2}\right\)^{-2}\cdot\left\(\frac{9}{4}\right\)) का मान क्या है?
What is the value of (\left\(\frac{3}{2}\right\)^{-2}\cdot\left\(\frac{9}{4}\right\))?
#polynomials
#fraction powers
#negative exponent
A (1)
B \(\frac{9}{4}\)
C \(\frac{4}{9}\)
D \(\frac{81}{16}\)
Explanation opens after your attempt
Step 1
Concept
(\left\(\frac{3}{2}\right\)^{-2}=\left\(\frac{2}{3}\right\)2 =\frac{4}{9}). Hence \(\frac{4}{9}\cdot\frac{9}{4}=1\).
Step 2
Why this answer is correct
The correct answer is A. (1). (\left\(\frac{3}{2}\right\)^{-2}=\left\(\frac{2}{3}\right\)2 =\frac{4}{9}). Hence \(\frac{4}{9}\cdot\frac{9}{4}=1\).
Step 3
Exam Tip
(\left\(\frac{3}{2}\right\)^{-2}=\left\(\frac{2}{3}\right\)2 =\frac{4}{9}) है। इसलिए \(\frac{4}{9}\cdot\frac{9}{4}=1\) है।
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(\left\(\frac{2}{3}\right\)0 +\left\(\frac{1}{2}\right\)2 ) का मान क्या है?
What is the value of (\left\(\frac{2}{3}\right\)0 +\left\(\frac{1}{2}\right\)2 )?
#polynomials
#zero exponent
#fraction power
A \(\frac{1}{4}\)
B \(\frac{3}{4}\)
C \(\frac{5}{4}\)
D \(\frac{9}{4}\)
Explanation opens after your attempt
Correct Answer
C. \(\frac{5}{4}\)
Step 1
Concept
Here (\left\(\frac{2}{3}\right\)0 =1) and (\left\(\frac{1}{2}\right\)2 =\frac{1}{4}). Therefore the sum is \(\frac{5}{4}\).
Step 2
Why this answer is correct
The correct answer is C. \(\frac{5}{4}\). Here (\left\(\frac{2}{3}\right\)0 =1) and (\left\(\frac{1}{2}\right\)2 =\frac{1}{4}). Therefore the sum is \(\frac{5}{4}\).
Step 3
Exam Tip
(\left\(\frac{2}{3}\right\)0 =1) और (\left\(\frac{1}{2}\right\)2 =\frac{1}{4}) है। इसलिए योग \(\frac{5}{4}\) है।
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(\left\(\sqrt{13}-\sqrt{5}\right\)\left\(\sqrt{13}+\sqrt{5}\right\)) का मान क्या है?
What is the value of (\left\(\sqrt{13}-\sqrt{5}\right\)\left\(\sqrt{13}+\sqrt{5}\right\))?
#real-numbers
#conjugate-product
#rational-result
A (8)
B (18)
C \(\sqrt{65}\)
D \(13+\sqrt{5}\)
Explanation opens after your attempt
Step 1
Concept
This is of the form ((a-b)(a+b)=a-2 -b-2 ).
Step 2
Why this answer is correct
(\(\sqrt{13}\)2 -\(\sqrt{5}\)2 =13-5=8).
Step 3
Exam Tip
In conjugate multiplication, directly use the difference of squares. चरण 1: यह ((a-b)(a+b)=a-2 -b-2 ) का रूप है। चरण 2: (\(\sqrt{13}\)2 -\(\sqrt{5}\)2 =13-5=8)। चरण 3: संयुग्म गुणन में वर्गों का अंतर सीधे लगाएं।
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(\left\(4+\sqrt{7}\right\)\left\(4-\sqrt{7}\right\)) का मान क्या है?
What is the value of (\left\(4+\sqrt{7}\right\)\left\(4-\sqrt{7}\right\))?
#real-numbers
#conjugates
#rational-result
A (9)
B (23)
C \(16+\sqrt{7}\)
D \(16-\sqrt{7}\)
Explanation opens after your attempt
Step 1
Concept
This is of the form ((a+b)(a-b)=a-2 -b-2 ).
Step 2
Why this answer is correct
(42 -\(\sqrt{7}\)2 =16-7=9).
Step 3
Exam Tip
In conjugate multiplication, directly use the difference of squares. चरण 1: यह ((a+b)(a-b)=a-2 -b-2 ) का रूप है। चरण 2: (42 -\(\sqrt{7}\)2 =16-7=9)। चरण 3: संयुग्म गुणन में वर्गों का अंतर सीधे लगाएं।
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(\left\(\sqrt{11}-\sqrt{2}\right\)\left\(\sqrt{11}+\sqrt{2}\right\)) का मान क्या है?
What is the value of (\left\(\sqrt{11}-\sqrt{2}\right\)\left\(\sqrt{11}+\sqrt{2}\right\))?
#real-numbers
#conjugate-product
#rational-result
A (9)
B (13)
C \(\sqrt{22}\)
D \(11+\sqrt{2}\)
Explanation opens after your attempt
Step 1
Concept
This is of the form ((a-b)(a+b)=a-2 -b-2 ).
Step 2
Why this answer is correct
(\(\sqrt{11}\)2 -\(\sqrt{2}\)2 =11-2=9).
Step 3
Exam Tip
In conjugate multiplication, directly use the difference of squares. चरण 1: यह ((a-b)(a+b)=a-2 -b-2 ) का रूप है। चरण 2: (\(\sqrt{11}\)2 -\(\sqrt{2}\)2 =11-2=9)। चरण 3: संयुग्म गुणन में वर्गों का अंतर सीधे लगाएं।
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(\left\(3+\sqrt{5}\right\)\left\(3-\sqrt{5}\right\)) का मान क्या है?
What is the value of (\left\(3+\sqrt{5}\right\)\left\(3-\sqrt{5}\right\))?
#real-numbers
#conjugates
#rational-result
A (4)
B (14)
C \(9+\sqrt{5}\)
D \(9-\sqrt{5}\)
Explanation opens after your attempt
Step 1
Concept
This is of the form ((a+b)(a-b)=a-2 -b-2 ).
Step 2
Why this answer is correct
(32 -\(\sqrt{5}\)2 =9-5=4).
Step 3
Exam Tip
In conjugate multiplication, directly use difference of squares. चरण 1: यह ((a+b)(a-b)=a-2 -b-2 ) का रूप है। चरण 2: (32 -\(\sqrt{5}\)2 =9-5=4)। चरण 3: संयुग्म गुणन में वर्गों का अंतर सीधे लगाएं।
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(\left\(\sqrt{7}-\sqrt{3}\right\)\left\(\sqrt{7}+\sqrt{3}\right\)) का मान क्या है?
What is the value of (\left\(\sqrt{7}-\sqrt{3}\right\)\left\(\sqrt{7}+\sqrt{3}\right\))?
#real-numbers
#conjugate-product
#rational-result
A (4)
B (10)
C \(\sqrt{21}\)
D \(7+\sqrt{3}\)
Explanation opens after your attempt
Step 1
Concept
This is of the form ((a-b)(a+b)=a-2 -b-2 ).
Step 2
Why this answer is correct
(\(\sqrt{7}\)2 -\(\sqrt{3}\)2 =7-3=4).
Step 3
Exam Tip
In conjugate multiplication, directly use the difference of squares. चरण 1: यह ((a-b)(a+b)=a-2 -b-2 ) का रूप है। चरण 2: (\(\sqrt{7}\)2 -\(\sqrt{3}\)2 =7-3=4)। चरण 3: संयुग्म गुणन में सीधे वर्गों का अंतर लगाएं।
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(\left\(2+\sqrt{3}\right\)\left\(2-\sqrt{3}\right\)) का मान क्या है?
What is the value of (\left\(2+\sqrt{3}\right\)\left\(2-\sqrt{3}\right\))?
#real-numbers
#conjugates
#rationalisation
A (1)
B (7)
C \(4+\sqrt{3}\)
D \(4-\sqrt{3}\)
Explanation opens after your attempt
Step 1
Concept
This is of the form ((a+b)(a-b)=a-2 -b-2 ).
Step 2
Why this answer is correct
(22 -\(\sqrt{3}\)2 =4-3=1).
Step 3
Exam Tip
For conjugate products, difference of squares gives the answer quickly. चरण 1: यह ((a+b)(a-b)=a-2 -b-2 ) का रूप है। चरण 2: (22 -\(\sqrt{3}\)2 =4-3=1)। चरण 3: संयुग्म रूप वाले गुणन में वर्गों का अंतर जल्दी उत्तर देता है।
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समीकरणों (7x+19y=86) और (13x+35y=158) में (a) और (b) के अनुपातों की तुलना से क्या निष्कर्ष निकलेगा?
What conclusion follows from comparing the ratios of (a) and (b) in the equations (7x+19y=86) and (13x+35y=158)?
#linear equations
#expert
#ratio comparison
#unique solution
A (7 / 13=19 / 35), इसलिए कोई हल नहीं / 35), so no solution
B (7 / 13=19 / 35), इसलिए अनंत हल / 35), so infinitely many solutions
C (7 / 13 \ne 19 / 35), इसलिए एक अद्वितीय हल / 35), so one unique solution
D तीनों अनुपात बराबर हैं / All three ratios are equal
Explanation opens after your attempt
Correct Answer
C. (7 / 13 \ne 19 / 35), इसलिए एक अद्वितीय हल / 35), so one unique solution
Step 1
Concept
The first two ratios are different. Therefore, the lines intersect at one point and give one unique solution.
Step 2
Why this answer is correct
The correct answer is C. \(7 / 13 \ne 19 / 35\), इसलिए एक अद्वितीय हल / 35), so one unique solution. The first two ratios are different. Therefore, the lines intersect at one point and give one unique solution.
Step 3
Exam Tip
पहले दो अनुपात अलग हैं। इसलिए रेखाएं एक बिंदु पर कटती हैं और एक अद्वितीय हल देती हैं।
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यदि किसी युग्म में गुणांक अनुपात समान है और स्थिर पद का अनुपात अलग है, तो हल-स्थिति क्या होगी?
If coefficient ratios are equal and the constant ratio is different in a pair, what is the solution status?
#class10
#linear-equations
#solvability
A अद्वितीय हल / Unique solution
B अनंत हल / Infinitely many solutions
C कोई हल नहीं / No solution
D सदैव दो हल / Always two solutions
Explanation opens after your attempt
Correct Answer
C. कोई हल नहीं / No solution
Step 1
Concept
This condition forms distinct parallel lines. Therefore, the pair is inconsistent.
Step 2
Why this answer is correct
The correct answer is C. कोई हल नहीं / No solution. This condition forms distinct parallel lines. Therefore, the pair is inconsistent.
Step 3
Exam Tip
यह स्थिति अलग-अलग समांतर रेखाएँ बनाती है। इसलिए युग्म असंगत होता है।
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समीकरणों (11x+18y=86) और (33x+54y=258) में तीनों अनुपातों का संबंध क्या है?
What is the relation among all three ratios in the equations (11x+18y=86) and (33x+54y=258)?
#linear equations
#expert
#ratio relation
#coincident
A तीनों बराबर हैं / All three are equal
B पहले दो बराबर और तीसरा अलग है / First two are equal and third is different
C पहले दो अलग हैं / First two are different
D केवल स्थिर पद बराबर हैं / Only constants are equal
Explanation opens after your attempt
Correct Answer
A. तीनों बराबर हैं / All three are equal
Step 1
Concept
Here (11/33=18/54=86/258). Therefore, both equations form the same line.
Step 2
Why this answer is correct
The correct answer is A. तीनों बराबर हैं / All three are equal. Here (11/33=18/54=86/258). Therefore, both equations form the same line.
Step 3
Exam Tip
यहां (11/33=18/54=86/258)। इसलिए दोनों समीकरण एक ही रेखा बनाते हैं।
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समीकरणों (19x+12y=71) और (9x+6y=35) में (a) और (b) के अनुपातों की तुलना से क्या पता चलता है?
What is found by comparing the ratios of (a) and (b) in the equations (19x+12y=71) and (9x+6y=35)?
#linear equations
#expert
#ratio comparison
#unique solution
A (19 / 9=12 / 6), इसलिए अनंत हल / 6), so infinitely many solutions
B (19 / 9=12 / 6), इसलिए कोई हल नहीं / 6), so no solution
C (19 / 9 \ne 12 / 6), इसलिए एक अद्वितीय हल / 6), so one unique solution
D (19 / 9=71 / 35), इसलिए संपाती / 35), so coincident
Explanation opens after your attempt
Correct Answer
C. (19 / 9 \ne 12 / 6), इसलिए एक अद्वितीय हल / 6), so one unique solution
Step 1
Concept
Here the first two ratios are different. Therefore, the lines intersect at one point and give one solution.
Step 2
Why this answer is correct
The correct answer is C. \(19 / 9 \ne 12 / 6\), इसलिए एक अद्वितीय हल / 6), so one unique solution. Here the first two ratios are different. Therefore, the lines intersect at one point and give one solution.
Step 3
Exam Tip
यहां पहले दो अनुपात अलग हैं। इसलिए रेखाएं एक बिंदु पर कटती हैं और एक हल देती हैं।
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समीकरणों (6x+17y=71) और (11x+31y=130) में (a) और (b) के अनुपातों की तुलना से क्या निष्कर्ष निकलेगा?
What conclusion follows from comparing the ratios of (a) and (b) in the equations (6x+17y=71) and (11x+31y=130)?
#linear equations
#hard
#ratio comparison
#unique solution
A (6 / 11=17 / 31), इसलिए कोई हल नहीं / 31), so no solution
B (6 / 11=17 / 31), इसलिए अनंत हल / 31), so infinitely many solutions
C (6 / 11 \ne 17 / 31), इसलिए एक अद्वितीय हल / 31), so one unique solution
D तीनों अनुपात बराबर हैं / All three ratios are equal
Explanation opens after your attempt
Correct Answer
C. (6 / 11 \ne 17 / 31), इसलिए एक अद्वितीय हल / 31), so one unique solution
Step 1
Concept
Here the first two ratios are different. Therefore, the lines intersect at one point and give one unique solution.
Step 2
Why this answer is correct
The correct answer is C. \(6 / 11 \ne 17 / 31\), इसलिए एक अद्वितीय हल / 31), so one unique solution. Here the first two ratios are different. Therefore, the lines intersect at one point and give one unique solution.
Step 3
Exam Tip
यहां पहले दो अनुपात अलग हैं। इसलिए रेखाएं एक बिंदु पर कटती हैं और एक अद्वितीय हल देती हैं।
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समीकरणों (9x+16y=74) और (27x+48y=222) में तीनों अनुपातों का संबंध क्या है?
What is the relation among all three ratios in the equations (9x+16y=74) and (27x+48y=222)?
#linear equations
#hard
#ratio relation
#coincident
A तीनों बराबर हैं / All three are equal
B पहले दो बराबर और तीसरा अलग है / First two are equal and third is different
C पहले दो अलग हैं / First two are different
D केवल स्थिर पद बराबर हैं / Only constants are equal
Explanation opens after your attempt
Correct Answer
A. तीनों बराबर हैं / All three are equal
Step 1
Concept
Here (9/27=16/48=74/222). Therefore, both equations form the same line.
Step 2
Why this answer is correct
The correct answer is A. तीनों बराबर हैं / All three are equal. Here (9/27=16/48=74/222). Therefore, both equations form the same line.
Step 3
Exam Tip
यहां (9/27=16/48=74/222)। इसलिए दोनों समीकरण एक ही रेखा बनाते हैं।
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समीकरणों (17x+10y=61) और (8x+5y=29) में (a) और (b) के अनुपातों की तुलना से क्या पता चलता है?
What is found by comparing the ratios of (a) and (b) in the equations (17x+10y=61) and (8x+5y=29)?
#linear equations
#hard
#ratio comparison
#unique solution
A (17 / 8=10 / 5) इसलिए अनंत हल / 5) so infinitely many solutions
B (17 / 8=10 / 5) इसलिए कोई हल नहीं / 5) so no solution
C (17 / 8 \ne 10 / 5) इसलिए एक अद्वितीय हल / 5) so one unique solution
D (17 / 8=61 / 29) इसलिए संपाती / 29) so coincident
Explanation opens after your attempt
Correct Answer
C. (17 / 8 \ne 10 / 5) इसलिए एक अद्वितीय हल / 5) so one unique solution
Step 1
Concept
Here the first two ratios are different. Therefore, the lines intersect at one point and give one solution.
Step 2
Why this answer is correct
The correct answer is C. \(17 / 8 \ne 10 / 5\) इसलिए एक अद्वितीय हल / 5) so one unique solution. Here the first two ratios are different. Therefore, the lines intersect at one point and give one solution.
Step 3
Exam Tip
यहां पहले दो अनुपात अलग हैं। इसलिए रेखाएं एक बिंदु पर कटती हैं और एक हल देती हैं।
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समीकरणों (7x+10y=39) और (5x+8y=31) में (a) और (b) के अनुपातों की तुलना से क्या निष्कर्ष निकलेगा?
What conclusion follows from comparing the ratios of (a) and (b) in the equations (7x+10y=39) and (5x+8y=31)?
#linear equations
#hard
#ratio comparison
#unique solution
A (7 / 5=10 / 8), इसलिए कोई हल नहीं / 8), so no solution
B (7 / 5=10 / 8), इसलिए अनंत हल / 8), so infinitely many solutions
C (7 / 5 \ne 10 / 8), इसलिए एक अद्वितीय हल / 8), so one unique solution
D तीनों अनुपात बराबर हैं / All three ratios are equal
Explanation opens after your attempt
Correct Answer
C. (7 / 5 \ne 10 / 8), इसलिए एक अद्वितीय हल / 8), so one unique solution
Step 1
Concept
Here the first two ratios are different. Therefore, the lines intersect at one point and give one unique solution.
Step 2
Why this answer is correct
The correct answer is C. \(7 / 5 \ne 10 / 8\), इसलिए एक अद्वितीय हल / 8), so one unique solution. Here the first two ratios are different. Therefore, the lines intersect at one point and give one unique solution.
Step 3
Exam Tip
यहां पहले दो अनुपात अलग हैं। इसलिए रेखाएं एक बिंदु पर कटती हैं और एक अद्वितीय हल देती हैं।
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समीकरणों (7x+10y=46) और (21x+30y=138) में तीनों अनुपातों का संबंध क्या है?
What is the relation among all three ratios in the equations (7x+10y=46) and (21x+30y=138)?
#linear equations
#hard
#ratio relation
#coincident
A तीनों बराबर हैं / All three are equal
B पहले दो बराबर और तीसरा अलग है / First two are equal and third is different
C पहले दो अलग हैं / First two are different
D केवल स्थिर पद बराबर हैं / Only constants are equal
Explanation opens after your attempt
Correct Answer
A. तीनों बराबर हैं / All three are equal
Step 1
Concept
Here (7/21=10/30=46/138). Therefore, both equations form the same line.
Step 2
Why this answer is correct
The correct answer is A. तीनों बराबर हैं / All three are equal. Here (7/21=10/30=46/138). Therefore, both equations form the same line.
Step 3
Exam Tip
यहां (7/21=10/30=46/138)। इसलिए दोनों समीकरण एक ही रेखा बनाते हैं।
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समीकरणों (13x+8y=47) और (6x+4y=23) में (a) और (b) के अनुपातों की तुलना से क्या पता चलता है?
What is found by comparing the ratios of (a) and (b) in the equations (13x+8y=47) and (6x+4y=23)?
#linear equations
#hard
#ratio comparison
#unique solution
A (13 / 6=8 / 4), इसलिए अनंत हल / 4), so infinitely many solutions
B (13 / 6=8 / 4), इसलिए कोई हल नहीं / 4), so no solution
C (13 / 6 \ne 8 / 4), इसलिए एक अद्वितीय हल / 4), so one unique solution
D (13 / 6=47 / 23), इसलिए संपाती / 23), so coincident
Explanation opens after your attempt
Correct Answer
C. (13 / 6 \ne 8 / 4), इसलिए एक अद्वितीय हल / 4), so one unique solution
Step 1
Concept
Here the first two ratios are different. Therefore, the lines intersect at one point and give one solution.
Step 2
Why this answer is correct
The correct answer is C. \(13 / 6 \ne 8 / 4\), इसलिए एक अद्वितीय हल / 4), so one unique solution. Here the first two ratios are different. Therefore, the lines intersect at one point and give one solution.
Step 3
Exam Tip
यहां पहले दो अनुपात अलग हैं। इसलिए रेखाएं एक बिंदु पर कटती हैं और एक हल देती हैं।
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समीकरणों (8x+12y=40) और (2x+3y=10) में तीनों अनुपातों का संबंध क्या है?
What is the relation among all three ratios in the equations (8x+12y=40) and (2x+3y=10)?
#linear equations
#hard
#ratio relation
#coincident
A तीनों बराबर हैं / All three are equal
B पहले दो बराबर और तीसरा अलग है / First two are equal and third is different
C पहले दो अलग हैं / First two are different
D केवल स्थिर पद बराबर हैं / Only constants are equal
Explanation opens after your attempt
Correct Answer
A. तीनों बराबर हैं / All three are equal
Step 1
Concept
Here (8/2=12/3=40/10). Therefore, both equations form the same line.
Step 2
Why this answer is correct
The correct answer is A. तीनों बराबर हैं / All three are equal. Here (8/2=12/3=40/10). Therefore, both equations form the same line.
Step 3
Exam Tip
यहां (8/2=12/3=40/10)। इसलिए दोनों समीकरण एक ही रेखा बनाते हैं।
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समीकरणों (5x+6y=32) और (15x+18y=96) में तीनों अनुपातों का संबंध क्या है?
What is the relation among all three ratios in the equations (5x+6y=32) and (15x+18y=96)?
#linear equations
#hard
#ratio relation
#coincident
A तीनों बराबर हैं / All three are equal
B पहले दो बराबर और तीसरा अलग है / First two are equal and third is different
C पहले दो अलग हैं / First two are different
D केवल स्थिर पद बराबर हैं / Only constants are equal
Explanation opens after your attempt
Correct Answer
A. तीनों बराबर हैं / All three are equal
Step 1
Concept
Here (5/15=6/18=32/96). Therefore, both equations form the same line.
Step 2
Why this answer is correct
The correct answer is A. तीनों बराबर हैं / All three are equal. Here (5/15=6/18=32/96). Therefore, both equations form the same line.
Step 3
Exam Tip
यहां (5/15=6/18=32/96)। इसलिए दोनों समीकरण एक ही रेखा बनाते हैं।
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समीकरणों (11x+6y=35) और (5x+3y=17) में (a) और (b) के अनुपातों की तुलना से क्या पता चलता है?
What is found by comparing the ratios of (a) and (b) in the equations (11x+6y=35) and (5x+3y=17)?
#linear equations
#hard
#ratio comparison
#unique solution
A (11 / 5=6 / 3) इसलिए अनंत हल / 3) so infinitely many solutions
B (11 / 5=6 / 3) इसलिए कोई हल नहीं / 3) so no solution
C (11 / 5 \ne 6 / 3) इसलिए एक अद्वितीय हल / 3) so one unique solution
D (11 / 5=35 / 17) इसलिए संपाती / 17) so coincident
Explanation opens after your attempt
Correct Answer
C. (11 / 5 \ne 6 / 3) इसलिए एक अद्वितीय हल / 3) so one unique solution
Step 1
Concept
Here the first two ratios are different. Therefore, the lines intersect at one point and give one solution.
Step 2
Why this answer is correct
The correct answer is C. \(11 / 5 \ne 6 / 3\) इसलिए एक अद्वितीय हल / 3) so one unique solution. Here the first two ratios are different. Therefore, the lines intersect at one point and give one solution.
Step 3
Exam Tip
यहां पहले दो अनुपात अलग हैं। इसलिए रेखाएं एक बिंदु पर कटती हैं और एक हल देती हैं।
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समीकरण (5x+6y=32) और (15x+18y=96) में तीनों अनुपातों का संबंध क्या है?
What is the relation among all three ratios in (5x+6y=32) and (15x+18y=96)?
#linear equations
#ratio relation
#coincident
A तीनों बराबर हैं / All three are equal
B पहले दो बराबर और तीसरा अलग है / First two are equal and third is different
C पहले दो अलग हैं / First two are different
D केवल स्थिर पद बराबर हैं / Only constants are equal
Explanation opens after your attempt
Correct Answer
A. तीनों बराबर हैं / All three are equal
Step 1
Concept
Here (5/15=6/18=32/96). Therefore both equations form the same line.
Step 2
Why this answer is correct
The correct answer is A. तीनों बराबर हैं / All three are equal. Here (5/15=6/18=32/96). Therefore both equations form the same line.
Step 3
Exam Tip
यहां (5/15=6/18=32/96)। इसलिए दोनों समीकरण एक ही रेखा बनाते हैं।
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समीकरण (11x+6y=35) और (5x+3y=17) में (a) और (b) के अनुपातों की तुलना से क्या पता चलता है?
What is found by comparing the ratios of (a) and (b) in (11x+6y=35) and (5x+3y=17)?
#linear equations
#ratio comparison
#unique solution
A (11 / 5=6 / 3) इसलिए अनंत हल / 3) so infinitely many solutions
B (11 / 5=6 / 3) इसलिए कोई हल नहीं / 3) so no solution
C (11 / 5 \ne 6 / 3) इसलिए एक अद्वितीय हल / 3) so one unique solution
D (11 / 5=35 / 17) इसलिए संपाती / 17) so coincident
Explanation opens after your attempt
Correct Answer
C. (11 / 5 \ne 6 / 3) इसलिए एक अद्वितीय हल / 3) so one unique solution
Step 1
Concept
Here the first two ratios are different. Therefore the lines intersect at one point and give one solution.
Step 2
Why this answer is correct
The correct answer is C. \(11 / 5 \ne 6 / 3\) इसलिए एक अद्वितीय हल / 3) so one unique solution. Here the first two ratios are different. Therefore the lines intersect at one point and give one solution.
Step 3
Exam Tip
यहां पहले दो अनुपात अलग हैं। इसलिए रेखाएं एक बिंदु पर कटती हैं और एक हल देती हैं।
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समीकरण (5x+7y=32) और (9x+13y=58) में (a) और (b) के अनुपातों की तुलना से क्या निष्कर्ष निकलेगा?
What conclusion follows from comparing the ratios of (a) and (b) in (5x+7y=32) and (9x+13y=58)?
#linear equations
#ratio comparison
#unique solution
A (5 / 9=7 / 13), इसलिए कोई हल नहीं / 13), so no solution
B (5 / 9 \ne 7 / 13), इसलिए एक अद्वितीय हल / 13), so one unique solution
C तीनों अनुपात बराबर हैं / All three ratios are equal
D रेखाएं संपाती हैं / Lines are coincident
Explanation opens after your attempt
Correct Answer
B. (5 / 9 \ne 7 / 13), इसलिए एक अद्वितीय हल / 13), so one unique solution
Step 1
Concept
Here \(5/9 \ne 7/13\), so the lines will intersect at one point. If the first two ratios differ, one unique solution is obtained.
Step 2
Why this answer is correct
The correct answer is B. \(5 / 9 \ne 7 / 13\), इसलिए एक अद्वितीय हल / 13), so one unique solution. Here \(5/9 \ne 7/13\), so the lines will intersect at one point. If the first two ratios differ, one unique solution is obtained.
Step 3
Exam Tip
यहां \(5/9 \ne 7/13\), इसलिए रेखाएं एक बिंदु पर कटेंगी। पहले दो अनुपात अलग हों तो एक अद्वितीय हल मिलता है।
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समीकरण (3x+4y=22) और (6x+8y=44) में तीनों अनुपातों का संबंध क्या है?
What is the relation among all three ratios in (3x+4y=22) and (6x+8y=44)?
#linear equations
#ratio relation
#coincident
A तीनों बराबर हैं / All three are equal
B पहले दो बराबर और तीसरा अलग है / First two are equal and third is different
C पहले दो अलग हैं / First two are different
D केवल स्थिर पद बराबर हैं / Only constants are equal
Explanation opens after your attempt
Correct Answer
A. तीनों बराबर हैं / All three are equal
Step 1
Concept
Here (3/6=4/8=22/44). Therefore, both equations form the same line.
Step 2
Why this answer is correct
The correct answer is A. तीनों बराबर हैं / All three are equal. Here (3/6=4/8=22/44). Therefore, both equations form the same line.
Step 3
Exam Tip
यहां (3/6=4/8=22/44)। इसलिए दोनों समीकरण एक ही रेखा बनाते हैं।
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समीकरण (9x+4y=21) और (4x+2y=10) में (a) और (b) के अनुपातों की तुलना से क्या पता चलता है?
What is found by comparing the ratios of (a) and (b) in (9x+4y=21) and (4x+2y=10)?
#linear equations
#ratio comparison
#unique solution
A (9 / 4=4 / 2), इसलिए अनंत हल / 2), so infinitely many solutions
B (9 / 4=4 / 2), इसलिए कोई हल नहीं / 2), so no solution
C (9 / 4 \ne 4 / 2), इसलिए एक अद्वितीय हल / 2), so one unique solution
D (9 / 4=21 / 10), इसलिए संपाती / 10), so coincident
Explanation opens after your attempt
Correct Answer
C. (9 / 4 \ne 4 / 2), इसलिए एक अद्वितीय हल / 2), so one unique solution
Step 1
Concept
Here the first two ratios are different. Therefore, the lines intersect at one point and give one solution.
Step 2
Why this answer is correct
The correct answer is C. \(9 / 4 \ne 4 / 2\), इसलिए एक अद्वितीय हल / 2), so one unique solution. Here the first two ratios are different. Therefore, the lines intersect at one point and give one solution.
Step 3
Exam Tip
यहां पहले दो अनुपात अलग हैं। इसलिए रेखाएं एक बिंदु पर कटती हैं और एक हल देती हैं।
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