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CAT 2022 Slot 3 — QA Question 22

3 Variable EquationsEasy

A donation box can receive only cheques of ₹100, ₹250, and ₹500. On one good day, the donation box was found to contain exactly 100 cheques amounting to a total sum of â‚¹15250. Then, the maximum possible number of cheques of ₹500 that the donation box may have contained, is 

Answer & solution

Answer: 12

Solution

Easy

Set up two equations — total count 100100 and total value Rs. 15250\text{Rs.}\,15250 — in the numbers of Rs.100100, Rs.250250, Rs.500500 cheques. Eliminate one variable to get a relation between the Rs.250250 and Rs.500500 counts, then maximise the latter.

1

Let a,b,ca,b,c be the counts of Rs.100100, Rs.250250, Rs.500500 cheques.

a+b+c=100(count)100a+250b+500c=15250 ⇒ 2a+5b+10c=305(value)\begin{aligned} &a+b+c=100 &&\text{(count)}\\ &100a+250b+500c=15250\ \Rightarrow\ 2a+5b+10c=305 &&\text{(value)} \end{aligned}
2

Eliminate aa: subtract 2×2\times(count) from (value):

(2a+5b+10c)−2(a+b+c)=305−200⇒ 3b+8c=105\begin{aligned} &(2a+5b+10c)-2(a+b+c)=305-200\\ &\Rightarrow\ 3b+8c=105 \end{aligned}
3

Maximise cc. From c=105−3b8c=\dfrac{105-3b}{8}, cc is largest when bb is smallest with cc an integer. We need 105−3b≡0(mod8)105-3b\equiv 0\pmod 8; the least valid bb is 33:

c=105−98=968=12c=\frac{105-9}{8}=\frac{96}{8}=12

(then a=100−3−12=85≥0a=100-3-12=85\ge 0, so it is feasible.)

Maximum number of Rs.500500 cheques is 12\mathbf{12}.

CAT 2022 Slot 3 QA Q22: A donation box can receive only cheques of ₹100, ₹250, and ₹500. On one good day, the donation box was f — Solution | TheCATExam