## Very important Maths questions for SSC CGL-13 Series#2

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## Very important Maths questions for SSC CGL-13 Series#2

21. Successive discount of 10% , 20% and 50% will be equivalent to a single discount of-

a. 36%

b. 64%

c. 80%

d. 56%

Ans : Equivalent to a single discount

= [100 – (100 – 10) (100 – 20) (100 – 50)/100*100]%

= [100 – 90*80*50/10000]%

= [100-36]%

= 64%

22. A retailer offers the following discount schemes for buyers on an article-

I. Two successive discounts of 10%

II. A discount of 12% followed by a discount of 8%.

III. Successive discounts of 15% and 5%

IV. A discount of 20%

The selling price will be minimum under the scheme-

a. I

b. II

c. III

d. IV

Ans : From (i) single discount = [10 + 10 – 10*10/100] % = 19%

From (ii) single discount = [12 + 8 -12*8/100] % = 19.04%

From (iii) single discount = [15 + 5 – 15*5/100] % = 19.25%

From (iv) single discount = 20%

We know The S.P. will be minimum under the scheme IV.

23. Of three numbers, the second is thrice the first and the third number is three-fourth of the first. If the average of the three numbers is 114, the largest number is –

a. 72

b. 216

c. 354

d. 726

Ans : Let the first number be x.

We know Second number = 3x and Third number = 3x/4

We know that x + 3x + 3x/4 = 3*114

=> 19x/4 = 342

We know x = 342*4/19 = 72

We know The largest number = 3*72

= 216

24. A car covers 1/5 of the distance from A to B at the speed of 8 km/hour, 1/10 of the distance at 25 km per hour and the remaining at the speed of 20 km per hour. Find the average speed of the whole journey-

a. 12.625 km/hr

b. 13.625 km/hr

c. 14.625 km/hr

d. 15.625 km/hr

Ans : If the whole journey be x km. The total time taken

= (x/5/8 + x/10/25 + 7x/10/20) hrs

= (x/40 + x/250 + 7x/200) hrs

= 25x + 4x + 35x/1000

= 64x/1000 hrs

We know Average speed = x/64x/1000

= 15.625 km/hr

25. The average of 3 numbers is 154. The first number is twice the second and the second number is twice the third. The first number is-

a. 264

b. 132

c. 88

d. 66

Ans : Let the first number be x.

We know Second number = x/2 and Third number = x/4

We know that x + x/2 + x/4 = 3 x 154

=> 7x/4 = 462

We know x = 462*4/7

=264

26. The average salary of all the staff in an office of a corporate house is Rs. 5,000. The average salary of the officers is Rs. 14,000 and that of the rest is Rs. 4,000. If the total number of staff is 500, the number of officers is –

a. 10

b. 15

c. 25

d. 50

Ans : Let the number of officers be x.

We know that 5000*500 = 14000x + 4000(500-x)

We know 2500000 =14000x + 2000000 – 4000x

We know x = 500000/10000

= 50

27. The average marks of 40 students in an English exam are 72. Later it is found that three marks 64, 62 and 84 were wrongly entered as 60, 65 and 73. The average after mistakes were rectified is-

a. 70

b. 72

c. 71.9

d. 72.1

Ans : Correct average

= 40*72 + (64 + 62 +84) – 68 – 65 – 73/40

= 2880 + 210 – 206/40 = 2884/40

= 72.1

28. A and B are two alloys of gold and copper prepared by mixing metals in the ratio 7:2 and 7:11 respectively. If equal quantities of the alloys are melted to form a third alloy C, the ratio of gold and copper in C will be-

a. 5 : 7

b. 5 : 9

c. 7 : 5

d. 9 : 5

Ans : Quantity of gold in A = 7/9* wt. of A

Quantity of gold in B = 7/18* wt. of B

If 1 kg of each A and B are mixed to form third alloys C.

Then quantity of gold in 2 kg C = 7/9 + 7/18

= 7/6 kg

And Quantity of copper in 2 kg C = 2 – 7/6

= 5/6 kg

We know Required ratio = 7/6 : 5/6 = 7 : 5

29. In a laboratory, two bottles contain mixture of acid and water in the ratio 2 : 5 in the first bottle and 7 : 3 in the second. The ratio in which the contents of these two bottles be mixed such that the new mixture has acid and water in the ratio 2 : 3 is-

a. 4 : 15

b. 9 : 8

c. 21 : 8

d. 1 : 2

Ans : Quantity of acid in first bottle = 2/7 x mix.

and Quantity of acid in second bottle = 7/10 x mix.

If x and 1 volumes are taken from I and II bottle respectively to form new mixture.

Then, (2/7 x + 7/10 * 1)/(5x/7 +3/10 * 1) = 2/3

=> 6x/7 21/10 = 10x/7 +6/10

=> 4x/7 = 15/10

We know x = 15/10*7/4 = 21/8

We know Required ratio = x : 1

= 21 : 8

30. A mixture contains 80% acid and rest water. Part of the mixture that should be removed and replaced by same amount of water to make the ratio of acid and water 4 : 3 is-

a. 1/3 rd

b. 3/7 th

c. 2/3 rd

d. 2/7 th

Ans : Let the initial wt. of mixture be 1 kg and x kg of mixture is taken out and replaced by same amount of water.

We know that Amt. of acid/Amt. of water = 0.8 – 0.8x/ (0.2 – 0.2x + x) = 4/3

=> 2.4 – 2.4x = 0.8 + 3.2x

=> 5.6x = 1.6

We know x = 1.6/5.6 = 2/7th part

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