Question 1
624
600
628
625
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Correct answer: Option II \frac{6}{10}R=\frac{2}{5}J R=\frac{2}{5}\times \frac{10}{6}J R=\frac{2}{3}J R/J = 2/3 R: J = 2:3 Therefore, Jitesh’s share = 2/5 *1500 = 600
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Question 2
7:9
7:2
6:5
8:3
Correct answer: Option I Solution: Let the third box be Z Then, first box = 140% of z = 140z/100 = 7/5 Then, second box = 180% of z = 180z/100 = 9/5 Therefore, ratios of the two boxes will be 7z/5 : 9z/5 = 35z : 45z = 7:9
Question 3
2000
3000
2894
1500
Solution: Let the shares be = 6x : 3x : 3x : 2x respectively Then 3x 2x = 1000 X = 1000 Therefore, Bharat’s share = 3x = 3 * 1000 = 3000
Question 4
4:5
8:15
2:5
2:17
Correct answer: Option II Solution: Let the number of seats left in the classes be = 4y and 6y Increased seats = (120/100* 4y) (150/100*6y) = 48y/10 , 9y = 48:90 The ratio will be 16: 30, = 8: 15
Question 5
30
23
75
Correct answer: Option I Solution: Quantity of juice = (30 *4/6) liters = 20 litres Quantity of water = (30 20) = 10 litres Another ratio = 2:4 Let the quantity of pulp to be added further be x times Then, juice: pulp (20/10+x) Now, (20/10+x) = 2/4 80 = 20 + 2x x = 60/2 = 30
Question 6
2:3
4:7
7:3
Correct answer: Option IV Explanation: Let the number of blue and red balls in the box be 4x and 8x respectively. Their increased number is (140% of 4x) and (120% of 8x). {140/100*4x} and {120/100*8x} The new ratio will be = 35x/5: 15x/5 = 7: 3
Question 7
15000
18000
20000
10000
Let the original salaries of Ronit and Sahil be Rs. 4x and Rs. 6x respectively
Question 8
11:1
11 : 22
10 : 1
12 : 1
Check formula used on this page here (Quick Method) S.P = 27.50 P = 25 % S.P = C.P x 125/100 27.50 = C.P x 125/100 C.P = 11 Rs. 24 0 \ / \ / 22 / \ / \ 22 2 W : S = 11:1 Ans (Long Solution)
Question 9
126, 230
740, 230
320, 480
231, 450
Solution: Let the number of coins be 4x and 6x. Total amount given = Rs.320 => (.25)(4x) + (.50)(6x) = 320 Hence we get x = 1x + 3x = 320 X = 80 = 25 paisa coins = 320 = 50 paisa coins = 480
Question 10
21
19
20
Suppose the can initially contains 7x and 5x of mixtures A and B respectively. Quantity of A in mixture left =7x  7/12 x 9 litres = 7x  21/4 litres. Quantity of B in mixture left = 5x  5/12 x 9 litres = 5x  15/4 litres. => 7x  21/4 / 5x  15/4 = 7/9 => 252x  189 = 140x + 147 => x = 3. i.e 7x = 7 x 3 So, the can contained 21 litres of A.
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March 26, 2021
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