**The difference of two numbers is 11 and one fifth of their sum is 9. The numbers are :**- 31,20
- 30,19
- 29,18
- 28,17

Answer : Option D.

x − y = 11, x + y = 5 × 9 x − y = 11, x + y = 45, y = 17, x = 28**How many numbers between 1 and 100 are divisible by 7 ?**- 9
- 11
- 17
- 14

Answer : Option D.

No. of divisible by 7 7, 14 --------- 98, n = a + (N - 1)d

98 = 7 + (N - 1) 7, 98 = 7 + 7N - 7

98/7= N = 14

**Three cubes of edges 6 cms, 8 cms and 10 cms are meted without loss of metal into a single cube. The edge of the new cube will be:**- 8 cms
- 12 cms
- 14 cms
- 16 cms

Answer : Option B.

Volume of new cube = Volume of cube 1 + cube 2 + cube 3 = 6^{3}+ 8^{3}+ 10^{3}, = 216 + 512 + 1000 a^{3}= 1728, a = (1728)^{1/3}= 12**If 378 coins consist of rupee, 50 paise and 25 paise coins, whose values are proportional to 13 :11 : 7, the number of 50 paise coins will be :**- 128
- 132
- 133
- 136

Answer : Option B.

If values are proportional to 13 : 11 : 7, then the number of coins will be proportional to 13/1 : 11/0.50 : 7/0.25 ⇒ 13 : 22 : 28. Now from this the number of coins of 50 paise will be 378 × 22/63 = 132.**In how many ways can 3 integers be selected from the set {1, 2, 3, …….., 37} such that sum of the three integers is an odd number?**- 3876
- 7638
- 6378
- 1938
- 969

Answer : Option A.

There are 18 even and 19 odd numbers in the given set. For sum to be odd either all 3 numbers should be odd or 2 of them even and one odd. This is possible in^{19}C_{3}+ (^{18}C_{2}×^{19}C_{1}) = 3876 waysSnapdeal Preparation Links- Sample Aptitude Questions of Snapdeal
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**The mean daily profit made by a shopkeeper in a month of 30 days was Rs. 350. If the mean profit for the first fifteen days was Rs. 275, then the mean profit for the last 15 days would be**- Rs. 200
- Rs. 350
- Rs. 275
- Rs. 425

Answer : Option D.

Average would be : 350 = (275 + x)/2 On solving, x = 425.**There were 35 students in a hostel. If the number of students increases by 7, the expenses of the mess increase by Rs. 42 per day while the average expenditure per head diminishes by Re 1. Find the original expenditure of the mess.**- Rs. 480
- Rs. 520
- Rs. 420
- Rs. 460

Answer : Option C.

Let d be the average daily expenditure

Original expenditure = 35 × d

New expenditure = 35 × d + 42

New average expenditure will be :

(35 × d + 42)/42 = d - 1

On solving, we get d = 12

Therefore original expenditure = 35 × 12 = 420**Six years ago, the ratio of the ages of Kunal and Sagar was 6 : 5. Four years hence, the ratio of their ages will be 11:10. What is Sagar's present age?**- 16 years
- 18 years
- 20 years
- 22 years
- 25 years

Answer : Option A.

Let the ages of Kunal and Sagar be K and S respectively.

Given : (K-6)/(S-6) = 6/5 => 5K – 6S = -6 ……(i)

And: (K+4)/(S+4) = 11/10 => 10K – 11 S = 4 …..(ii)

Solving (i) & (ii) we get: S = 16 years.**A number X is 150 more than a second number, Y. If the sum of X and Y is 5 times Y, what is the value of Y?**- 50
- 40
- 80
- 60
- 70

Answer : Option A.

X = 150 + Y

X + Y = 5Y

150 + 2Y = 5Y

150 = 3Y => Y = 50**A square field has an area of 50625 m2. Find the cost of fencing around it at Rs. 15 per metre, (in Rs.)**- 12,500
- 6750
- 17,500
- 13,500
- 16,250

Answer : Option D.

Area of square field = 50625 => side of the field = √50625 = 225 cost of fencing @ Rs. 15/ meter = 4 × 225 × 15 = Rs. 13500