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No. of Questions 
10 
Time 
10 min 
Medium 
english 
Marks 
10 
Positive Marks 
+1 
Negative Marks 
0 
INSTRUCTION : All question carry 1 mark and there is only single option correct for every questionMark review to see the question later 
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Question 1 of 10
1. Question
1 pointsSolve the equation for x : 6x – 27 + 3x = 4 + 9 – x
Correct
Answer: Option A
Explanation:
19x + 19y + 17 = 19x + 19y – 21
38x = 38 => x = 1Incorrect
Answer: Option A
Explanation:
19x + 19y + 17 = 19x + 19y – 21
38x = 38 => x = 1 
Question 2 of 10
2. Question
1 pointsThe cost of 2 chairs and 3 tables is Rs.1300. The cost of 3 chairs and 2 tables is Rs.1200. The cost of each table is more than that of each chair by?
Correct
Answer: Option E
Explanation:
2C + 3T = 1300 — (1)
3C + 3T = 1200 — (2)
Subtracting 2nd from 1st, we get
C + T = 100 => T – C = 100Incorrect
Answer: Option E
Explanation:
2C + 3T = 1300 — (1)
3C + 3T = 1200 — (2)
Subtracting 2nd from 1st, we get
C + T = 100 => T – C = 100 
Question 3 of 10
3. Question
1 pointsThe denominator of a fraction is 1 less than twice the numerator. If the numerator and denominator are both increased by 1, the fraction becomes 3/5. Find the fraction?
Correct
Answer: Option D
Explanation:
Let the numerator and denominator of the fraction be ‘n’ and ‘d’ respectively.
d = 2n – 1
(n + 1)/(d + 1) = 3/5
5n + 5 = 3d + 3
5n + 5 = 3(2n – 1) + 3 => n = 5
d = 2n – 1 => d = 9
Hence the fraction is : 5/9.Incorrect
Answer: Option D
Explanation:
Let the numerator and denominator of the fraction be ‘n’ and ‘d’ respectively.
d = 2n – 1
(n + 1)/(d + 1) = 3/5
5n + 5 = 3d + 3
5n + 5 = 3(2n – 1) + 3 => n = 5
d = 2n – 1 => d = 9
Hence the fraction is : 5/9. 
Question 4 of 10
4. Question
1 points5. The cost of 10 kg of apples is equal to the cost of 24 kg of rice. The cost of 6 kg of flour equals the cost of 2 kg of rice. The cost of each kg of flour is Rs.20.50. Find the total cost of 4 kg of apples, 3 kg of rice and 5 kg of flour?
Correct
Answer: Option B
Explanation:
Let the costs of each kg of apples and each kg of rice be Rs.a and Rs.r respectively.
10a = 24r and 6 * 20.50 = 2r
a = 12/5 r and r = 61.5
a = 147.6
Required total cost = 4 * 147.6 + 3 * 61.5 + 5 * 20.5
= 590.4 + 184.5 + 102.5 = Rs.877.40Incorrect
Answer: Option B
Explanation:
Let the costs of each kg of apples and each kg of rice be Rs.a and Rs.r respectively.
10a = 24r and 6 * 20.50 = 2r
a = 12/5 r and r = 61.5
a = 147.6
Required total cost = 4 * 147.6 + 3 * 61.5 + 5 * 20.5
= 590.4 + 184.5 + 102.5 = Rs.877.40 
Question 5 of 10
5. Question
1 pointsFind the roots of the quadratic equation: x^{2} + 2x – 15 = 0?
Correct
Answer: Option A
Explanation:
x^{2} + 5x – 3x – 15 = 0
x(x + 5) – 3(x + 5) = 0
(x – 3)(x + 5) = 0
=> x = 3 or x = 5.Incorrect
Answer: Option A
Explanation:
x^{2} + 5x – 3x – 15 = 0
x(x + 5) – 3(x + 5) = 0
(x – 3)(x + 5) = 0
=> x = 3 or x = 5. 
Question 6 of 10
6. Question
1 pointsFind the roots of the quadratic equation: 2x^{2} + 3x – 9 = 0?
Correct
Answer: Option B
Explanation:
2x^{2} + 6x – 3x – 9 = 0
2x(x + 3) – 3(x + 3) = 0
(x + 3)(2x – 3) = 0
=> x = 3 or x = 3/2.Incorrect
Answer: Option B
Explanation:
2x^{2} + 6x – 3x – 9 = 0
2x(x + 3) – 3(x + 3) = 0
(x + 3)(2x – 3) = 0
=> x = 3 or x = 3/2. 
Question 7 of 10
7. Question
1 pointsThe roots of the equation 3x^{2} – 12x + 10 = 0 are?
Correct
Answer: Option D
Explanation:
The discriminant of the quadratic equation is (12)^{2} – 4(3)(10) i.e., 24. As this is positive but not a perfect square, the roots are irrational and unequal.
Incorrect
Answer: Option D
Explanation:
The discriminant of the quadratic equation is (12)^{2} – 4(3)(10) i.e., 24. As this is positive but not a perfect square, the roots are irrational and unequal.

Question 8 of 10
8. Question
1 pointsIf the roots of a quadratic equation are 20 and 7, then find the equation?
Correct
Answer: Option C
Explanation:
Any quadratic equation is of the form
x^{2} – (sum of the roots)x + (product of the roots) = 0 — (1)
where x is a real variable. As sum of the roots is 13 and product of the roots is 140, the quadratic equation with roots as 20 and 7 is: x^{2} – 13x – 140 = 0.Incorrect
Answer: Option C
Explanation:
Any quadratic equation is of the form
x^{2} – (sum of the roots)x + (product of the roots) = 0 — (1)
where x is a real variable. As sum of the roots is 13 and product of the roots is 140, the quadratic equation with roots as 20 and 7 is: x^{2} – 13x – 140 = 0. 
Question 9 of 10
9. Question
1 pointsThe sum and the product of the roots of the quadratic equation x^{2} + 20x + 3 = 0 are?
Correct
Answer: Option E
Explanation:
Sum of the roots and the product of the roots are 20 and 3 respectively.
Incorrect
Answer: Option E
Explanation:
Sum of the roots and the product of the roots are 20 and 3 respectively.

Question 10 of 10
10. Question
1 pointsIf the roots of the equation 2x^{2} – 5x + b = 0 are in the ratio of 2:3, then find the value of b?
Correct
Answer: Option A
Explanation:
Let the roots of the equation 2a and 3a respectively.
2a + 3a = 5a = ( 5/2) = 5/2 => a = 1/2
Product of the roots: 6a^{2} = b/2 => b = 12a^{2}
a = 1/2, b = 3.Incorrect
Answer: Option A
Explanation:
Let the roots of the equation 2a and 3a respectively.
2a + 3a = 5a = ( 5/2) = 5/2 => a = 1/2
Product of the roots: 6a^{2} = b/2 => b = 12a^{2}
a = 1/2, b = 3.