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Linear Equations

Linear Equations

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Basics of Equations
Graph, Dependent & Inconsistent Equations
Solving Equations
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Integer Solutions
2 Equations & 3 Variables
Items Measured in Groups of 2 or more
Transposed & Multiple Variables
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Linear Equations 1
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Linear Equations 2
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Linear Equations : Level 1
Linear Equations : Level 2
Linear Equations : Level 3
ALL MODULES

CAT 2025 Lesson : Linear Equations - Transposed & Multiple Variables

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4.9 Interchanged/Transposed

Example 19

A man received a cheque. The amount in Rs. has been transposed for paise and vice versa. After spending Rs. 555 and 424242 paise, he discovered he now had exactly 666 times the value of the correct cheque amount. What amount he should have received? [CAT 2002] (111) 5.305.305.30            (222) 6.446.446.44            (333) 60.4460.4460.44            (444) 16.4416.4416.44           

Solution

Let
xxx and yyy be the amount rupees and paise that he should ideally have received respectively. Amount he should have received in paise =100x+y= 100x + y=100x+y

Amount in paise, after transposing
=100y+x= 100y + x=100y+x

After spending
542542542 paise, he had 666 times the value of the correct cheque amount.
⇒
100y+x−542=6(100x+y) 100y + x - 542 = 6(100x + y)100y+x−542=6(100x+y)
⇒
94y=599x+542 94y = 599x + 54294y=599x+542

⇒
y=599x+54294 y = \dfrac{599x + 542}{94}y=94599x+542​

We divide
599599599 and 542542542 by 949494, write the quotients separately and then the remainders.

⇒
y=6x+5+35x+7294 y = 6x + 5 + \dfrac{35x + 72}{94}y=6x+5+9435x+72​

Substituting values for
xxx from 111 onwards, we note that x=6x = 6x=6 results in an integer for yyy at y=44y = 44y=44. Therefore, the amount is Rs. 6.44\bm{6.44}6.44.

Alternatively (Recommended Method)

As this question has options, it is best to run through the options. Options
333 and 444 as original numbers are very high. This would have resulted in the transpose becoming a 333-digit number.

Option
(1)\bm{(1)}(1): 5.30×6+5.42=31.80+5.42=37.225.30 \times 6 + 5.42 = 31.80 + 5.42 = 37.225.30×6+5.42=31.80+5.42=37.22
Option
(2)\bm{(2)}(2): 6.44×6+5.42=38.64+5.42=44.066.44 \times 6 + 5.42 = 38.64 + 5.42 = 44.066.44×6+5.42=38.64+5.42=44.06
In option
222 the rupees and paise of Rs. 06.4406.4406.44 are transposed in the cheque as 44.0644.0644.06.

Answer: (
222) 6.446.446.44


4.10 Multiple Variables

Example 20

Mr. and Mrs. Gupta have three children – Pratik, Writtik and Kajol, all of whom were born in different cities. Pratik is 222 years elder to Writtik. Mr. Gupta was 303030 years of age when Kajol was born in Hyderabad, while Mrs. Gupta was 282828 years of age when Writtik was born in Bangalore. If Kajol was 555 years of age when Pratik was born in Mumbai, then what were the ages of Mr. and Mrs. Gupta respectively at the time of Pratik’s birth? [IIFT 2011] (111) 353535 years, 262626 years       
(
222) 303030 years, 212121 years       
(
333) 373737 years, 282828 years       
(
444) None of the above           

Solution

Let
a,b,p,wa, b, p, wa,b,p,w and kkk be the ages (in years) of Mr. Gupta, Mrs. Gupta, Pratik, Writtik and Kajol respectively.

The following equations are formed from the data in the question.
p=w+2p = w + 2p=w+2 -----(111)
a=k+30a = k + 30a=k+30 -----(222)
b=w+28b = w + 28b=w+28 -----(333)
k=p+5k = p + 5k=p+5 -----(444)

We need to find Mr. and Mrs. Gupta's age at the time of Pratik's birth, which is nothing but the difference between their respective ages and that of Pratik. Therefore, we need to find
a−pa - pa−p and b−pb - pb−p.

Equations (
222) and (444) have kkk as the common variable and also contains aaa and ppp.
∴\therefore∴ Substituting Eq(444) in Eq(242424) ⇒ a=p+5+30 a = p + 5 + 30a=p+5+30 ⇒ a−p=35a - p = \bm{35}a−p=35

Likewise, substituting
w=p−2w = p - 2w=p−2 from Eq(111) in Eq(333) ⇒ b=p−2+28 b = p - 2 + 28b=p−2+28 ⇒ b−p=26b - p = \bm{26}b−p=26

Mr. Gupta and Mrs. Gupta would have been
35\bm{35}35 years and 26\bm{26}26 years respectively when Pratik was born.

Answer: (
111) 353535 years, 262626 years


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