# NUMBERS & FUNDAMENTALS OF MATHS (PART-VII)

#### QUERY 61

**A man bought two baskets full of mangoes for Rs 19.80 and Rs 34.65. Cost of each mango is not less than 24 paise and not greater than 36 paise. The total number of mangoes bought by him is?**

A) 135

B) 145

C) 155

D) 165

**MAHA GUPTA**

Do it by options. The total cost of the mangoes must be divisible by the total number of mangoes bought.

Now total cost = 19.80 + 34.65 = Rs = Rs 54.45 = 5445 Paise

As unit digit in the cost price and all the options is 5, so check the divisibility only by the other factors of the options.

Now the factors

135 = 5*27

145 = 5*29

155 = 5*31

165 = 5*3*11

Obviously we’ll check it first for the last option as all the factors are relatively smaller in size and co-prime.

We see this divides the cost price completely; hence the answer (option ‘D’)

#### QUERY 62

**The cost of 7 shirts, 10 jackets and 13 trousers is Rs 25000 whereas cost of 25 trousers, 17 shirts and 21 jackets is Rs 41000. Find the cost of 1 shirt, 1 jacket, 1 trouser combined.**

A) Rs 900

B) Rs 1500

C) Rs 1200

D) Rs 1000

**SHIV KISHOR**

Let cost of one jacket is J, one shirt is S, and one trouser is T

Also let J + S + T = x . Obviously we have to find the value of x.

Now, 7S + 10J +13T = 25000 …….(i)

and 17S + 21J + 25T = 41000 ……..(ii)

On subtracting (i) from (ii)

10S + 11J + 12T = 16000

=> 10(S+J+T) + J + 2T = 16000

=> J + 2T = 16000 – 10x (putting J + S + T = x)

=> 3J + 6T = 48000 – 30x

Now from (i)

7(J+S+T) + 3J + 6T = 25000

=> 7x + 48000 – 30x = 25000 (putting 3J + 6T = 48000 – 30x)

=> -23x = -23000

=> x = 1000 (option ‘D’)