GMAT Linear Equations Questions With Solutions

1 rupee spent on a carrot gives 1.66 units of Vitamin A, while 1 rupee spent on a beetroot gives 1.5 units of Vitamin A.
So, 40 carrots and 0 beetroots give the maximum amount of Vitamin A

Let n1 and n2 be the two numbers given
$$2/5 n_1 \times 1/7 n_2=16 \times n_2$$
∴ $$n_1 = 280$$ but $$n_2$$ cannot be determined. Choice (d)

Let the limit be x and the rate of charge be k per kg.
Let the excess luggage with Raja be R kg.
So, excess luggage with Praja = 2R kg
Now, excess luggage with Raja + excess luggage with Praja = 60 - 2x
So, 3R = 60 - 2x => R = 20 - 2x/3 which was charged 1200 Also, if one person had the entire luggage, excess luggage would have been 60 - x, which would have been charged 5400.
=> (60-2x)/3*(60-x) = 1200/5400
Solving this, x = 15 kg
So, Praja's luggage = 35 kg

Let the limit be x and the rate of charge be k per kg.
Let the excess luggage with Raja be R kg.
So, excess luggage with Praja = 2R kg
Now, excess luggage with Raja + excess luggage with Praja = 60 - 2x
So, 3R = 60 - 2x => R = 20 - 2x/3 which was charged 1200 Also, if one person had the entire luggage, excess luggage would have been 60 - x, which would have been charged 5400.
=> (60-2x)/3*(60-x) = 1200/5400
Solving this, x = 15 kg

Let the number be xy
10y + x = 10x + y + 18
=> 9y - 9x = 18
=> y - x = 2
So, y can take values from 9 to 4 (since 3 is already counted in 13)
Number of possible values = 6

Let Rita's age be R.
Mother's age = 3R/2
Daughter's age = R/4
Total = R + 3R/2 + R/4 = 11R/4 = 154 => R = 56
So, difference of ages = R/2 = 28 years

2x2 + 3x - 20 = 0
x = -8 , 5
2y2 + 19y + 44 = 0
y = -11 , -8

x ≥ y

10x2 - 7x + 1 = 0
x = 1/5 , 1/2
35y2 - 12y + 1 = 0
y = 1/7 , 1/5
x >= y

If one eighths of the number is 25.5, the number is 25.5*8 = 204. One sixths of 204 = 34

$$\frac{?}{432}=\frac{243}{?}$$
Let the unknown be x
x^2=243*432
x=324
Option C is correct.

Now, (5/12) of 516 = (5/12)*516 = 215
(4/9) of 495 = 220
215 is less than 220 by 5.
Hence, correct option is E.

Let N be total number of sweets and each student get x sweet.
N/200 = x, N=200x
Also, as per given condition,
N/160 = x+2
N = 160x + 320
200x = 160x + 320
40x = 320
x = 8
Number of sweets = 1600
Hence, the correct option is option D.

Let the price of a digital camera be x and the price of a handy camera be y.

So, 15x + 21y = 354900

5x + 7y = (15x + 21y)/3 = 354900 / 3 = 118300

56 kg rice per week => 8 kg rice per day

Number of days in April + May = 61

So, quantity of rice required = 61 * 8 = 488 kg

Amount he wanted to withdraw = Rs. 49350

Amount he withdrew = Rs. 48150

So amount that will be credited back to his account = Rs. 49350 - Rs. 48150 = Rs. 1200

.After adding Rs. 1200 to Rs. 25376, the current balance the new balance will be Rs. 25376 + Rs. 1200 = Rs. 26576

Hence Option A.

Let the cost of a shirt be S and the cost of a pant be P.
So, 18S+21P = 42
Multiplying the equation by 5, we get 90S+105P=210
Hence, the cost of 90 shirts and 105 pants equals Rs. 210

Let the cost of a pencil be A and let the cost of a pen be B.
So, 15B+20A = 200
Multiplying the above equation by 12, we get 180B+240A = 2400
Hence, the cost of 180 pens and 240 pencils is Rs. 2400

Cost of 12 belts + cost of 30 wallets = Rs 8940

Cost of $$\frac{1}{3}\times$$12 belts + cost of $$\frac{1}{3}\times30$$wallets

= $$ \frac{1}{3}\times8940$$ = Rs. 2980

Let their ages today be 5x and 4x. Their ages 10 years from now are: 5x + 10 and 4x + 10. But 5x + 10: 4x + 10 = 6 : 5
So, 25x + 50 = 24x + 60
So, x = 10
Or Ravi’s present age is 50 years.

The cost of 30 calculators and 36 watches is three times the cost of 10 calculators and 12 watches.
Hence, the required cost equals 11100*3 = Rs. 33,300

Cost of each bat be 'x'
Cost of each stick be 'y'
Given that, $$24x + 32y = 5600$$
$$8 \times(3x + 4y) = 5600$$
$$3x + 4y = 700$$
Hence, cost of 3 bats and 4 sticks is Rs. 700

Let the cost of 1 chocolate be x and the cost of 1 packet biscuit be y.
6x+8y = 200
3x+4y = 100
Multiply both sides by 5
15x+20y = 5*100 = 500
Hence the cost of 15 chocolates and 20 packets of biscuits is Rs 500.

Let the age of Arun be x. Then, the age of Shyam is 3x and the age of Ram is 3x + 5.
Sum of their ages = x + 3x + 3x + 5 = 7x + 5 = 40
=> x = 5.
Thus, age of Ram = 3x + 5 = 20.

Let the first number be x and the second number be y.
Given, 6/11 of x = 60/100 of y
=> x/y = 11/10

Let x be the lowest of the seven consecutive integers.
The sum of the seven integers = x + x + 1 + x + 2 + x + 3 + x + 4 + x + 5 + x + 6 = 7x + 21 = 52x7
7x = 364 - 21
x = 343/7 = 49
Thus, the middle integer = x + 3 = 49 + 3 = 52.

5x + 6y = 17
3x + 8y = 19

Multiplying the first equation by 3 and the second equation by 5, we get,

15x + 18y = 51
15x + 40y = 95

=> 22y = 44 => y = 2
Solving for x, we get x = 1
So, x < y. Option b) is the correct answer.

Let the two numbers be x and y.
Hence, 55% of x = 3y/8
=> 55x/100 = 3y/8
11x/20 = 3y/8
x/y = (20 * 3)/(8*11) = 15/22

$$4^{a + b} = 1024 = 4^{5} => a + b = 5 (1)
$$4^{a - b} = 4$$ => a - b = 1 (2)
On solving (1) and (2) we get a = 3, b = 2.
Thus, ab = 3 x 2 = 6.

1985 - 3x = 4x + 445
=> 7x = 1540
=> x = 220

Let the cost of apples be x
Let the cost of oranges be y.

Given,
3x + 2y = 65 … i

2x + 5y = 80 …. ii

Multiplying i by 2 and ii by 3, and subtracting the equations,

11y = 110
y = 10

substituting y in i, x = 15

Difference between 10 apples and 8 oranges = 10x - 8y = 150 - 80 = 70

All the letters are distinct. Hence, the number of ways in which the letters of the word VIRTUAL can be arranged = 7! = 5040.

6 days $$\rightarrow$$ 27 bundles.
1 days $$\rightarrow$$ 4.5 bundles.
14 days $$\rightarrow$$ 63 bundles.
Hence, Option C is correct.

324 pencils$$\rightarrow$$6years.
324/6 pencils$$\rightarrow$$1year.
$$54\times14$$pencils$$\rightarrow$$ 14years.
Hence, students use 756 pencils in 14 years.
Therefore, No. of dozens of pencils=$$\frac{756}{12}$$.
=$$63$$.
Hence, Option E is correct.

Let the number = $$x$$

Acc to ques,

=> $$\frac{3x}{5} - 33.75 = \frac{45}{100} x$$

=> $$\frac{3x}{5} - \frac{9x}{20} = 33.75$$

=> $$\frac{3x}{20} = 33.75$$

=> $$x = \frac{33.75 \times 20}{3} = 225$$

$$\therefore$$ One-third of the number = $$\frac{1}{3} \times 225$$

= 75

Let price of shirt = Rs. $$x$$

Price of trouser = Rs. $$y$$

=> $$5x + 6y = 2340$$ -----------Eqn(1)

and $$7x - 3y = 540$$ -----------Eqn(2)

Multiplying eqn(2) by 2 and adding both equations, we get :

=> $$19x = 3420$$

=> $$x = \frac{3420}{19} = 180$$

$$\therefore$$ Cost of 4 shirts = 180 * 4 = Rs. 720

Let the cost of a fan = Rs. $$x$$

Cost of an oven = Rs. $$y$$

=> $$8x + 14y = 36520$$

Multiplying both sides by $$(\frac{3}{2})$$, we get :

=> $$\frac{3}{2} \times (8x + 14y = 36520)$$

=> $$12x + 21y = 54,780$$

Cost of 36 pens and 42 pencils = 460

Dividing by 2, we get :

=> cost of 18 pens and 21 pencils = $$\frac{460}{2}$$ = Rs. 230

C.P. of 1 pen = Rs. $$x$$

C.P. of 1 pencil = Rs. $$y$$

=> $$20x + 17y = 418$$

Multiplying above equations by 3

$$\therefore$$ $$60x + 51y = 1254$$

Case I : Trader sells 150 metres of cloth for Rs. 6,600

=> S.P. per metre = Rs. $$\frac{6600}{150}$$

= Rs. 44

Case II : Trader sells 300 metres of cloth for Rs. 12,750

=> S.P. per metre = Rs. $$\frac{12750}{300}$$

= Rs. 42.5

$$\therefore$$ Required discount = Rs. (44 - 42.5) = Rs. 1.5

Let the cost of a folder = Rs. $$x$$

and cost of a pen = Rs. $$y$$

=> $$20x + 15y = 995$$

Multiplying the above equation by $$(\frac{3}{5})$$, we get :

=> $$\frac{3}{5} \times (20x + 15y = 995)$$

=> $$12x + 9y = 597$$

$$\therefore$$ Cost of 12 folders and 9 pens is Rs. 597

There are 30 students and 2 teachers.

Sweets received by each student = $$\frac{20}{100} * 30$$ = 6

=> Total sweets received by all the students = 30 * 6 = 180

Sweets received by each teacher = $$\frac{30}{100} * 30$$ = 9

=> Total sweets received by both teachers = 2 * 9 = 18

$$\therefore$$ Total sweets distributed in class = 180 + 18 = 198

Total amount = 97836

No. of children = 31

Since amount is distributed equally

=> Amount each child will get = $$\frac{97836}{31}$$ = Rs. 3,156

Let us rank the person according to the marks scored by them, where 1 -> highest marks and 5 -> lowest marks.

Only one person scored less marks than B, => B = 4

Also, A > D > B and $$A \neq 1$$

=> A = 2 and D = 3

Thus, ranking from 1-5 = C/E , A , D , B , E/C

$$\therefore$$ A scored the second highest marks.

Let P has = $$Rs. 100x$$

=> Amount with Q = $$100x + \frac{25}{100} \times 100x = Rs. 125x$$

=> Amount with R = $$\frac{1}{5} \times 125x = Rs. 25x$$

Total amount together = $$100x + 125x + 25x = 150$$

=> $$x = \frac{150}{250} = \frac{3}{5}$$

=> $$x = 0.6$$

$$\therefore$$ Amount with P alone = $$100 \times 0.6 = Rs. 60$$

$$\dfrac{49}{a + b} + \dfrac{81}{a - b} = 34$$ Eq.(1)

$$\dfrac{84}{a - b} + \dfrac{245}{a + b} = 63$$ Eq.(2)

Let’s assume $$\dfrac{1}{a + b} = x$$ and $$\dfrac{1}{a - b} = y$$.

Now put these in Eq.(1) and Eq.(2).
$$49x + 81y = 34$$ Eq.(3)
$$84y + 245x = 63$$
$$245x + 84y = 63$$ Eq.(4)
Multiply Eq.(3) by 5.
$$245x + 405y = 170$$ Eq.(5)
Subtract Eq.(4) from Eq.(5).
245x + 405y - (245x + 84y) = 170 - 63
245x + 405y - 245x - 84y = 170 - 63
321y = 107

$$y = \dfrac{107}{321}$$

$$y = \dfrac{1}{3}$$ Eq.(6)

Put Eq.(6) in Eq.(3).

$$49x + 81y = 34$$

$$49x + 81\times\dfrac{1}{3} = 34$$

49x + 27 = 34
49x = 34 - 27
49x = 7

$$x = \dfrac{7}{49}$$

$$x = \dfrac{1}{7}$$ Eq.(7)

As we assumed initially that $$\dfrac{1}{a + b} = x$$ and $$\dfrac{1}{a - b} = y$$.

Put Eq.(6) and Eq.(7) here to obtain the value of a and b.

$$\dfrac{1}{a + b} = \dfrac{1}{7}$$

a + b = 7 Eq.(8)

Similarly $$\dfrac{1}{a - b} = \dfrac{1}{3}$$

a - b = 3 Eq.(9)
Add Eq.(8) and Eq.(9).
2a = 7 + 3
2a = 10
a = 5 Eq.(10)
Put Eq.(10) in Eq.(9).
5 - b = 3
b = 5 - 3 = 2 Eq.(11)

Now compare Eq.(10) and Eq.(11).
5 > 2
a > b
Hence, option e is the correct answer.