GMAT Data Sufficency Questions With Solutions, Download PDF

Statement A:
y - x = 20
∴ we cannot find the value of p.

Statement B:
X + y = 1620
∴ we can’t find the value of p.

Statement C: P(1+2r/100)= 9,600

Using A and B together,
X = 800, y = 820
∴ On 400, the interest accrued is 8000.
∴ Capital P = 400 * 100/5

A and B together sufficient.
A and C together (or) B and C together are not sufficient. Choice a

A. Total marks in 4 subjects including English
=4×60=240
B. Total marks in English and Maths =170
C. Total marks in Maths and Science=180
The question can’t be answered because nothing has been said about the marks in the fourth subject.
Also, there are four unknowns but only three questions can be formed with give data.

Length of the train = 240 m
Time taken to cross the pole = 24 seconds
Speed = $$\frac{240}{24}$$ = 10 m/s

SI in 1 year = 600

for two years = 1200

Sum = 10 x 1200 = 12000

600 = 12000*r*1/100

r = 5%

Amount returned in two years = 12000(1 + $$\frac{5}{100}$$)2 = 13230

Using options one and two

By first statement,

Difference between CI and SI for two years = $$P(1+R)^2$$ - P - 2PR

= $$PR^2$$ = 100

By second statement,

P + 3PR = 19500

Taking the ratio of both $$R^2$$ / (1 + 3R ) = 100/19500 = 1/195

=> $$195R^2 - 3R -1 = 0$$

=> R $$\approx$$ 8% or - 6.4%. As R cannot be negative it will be 8%.

Using options one and three,

$$PR^2$$ = 100

2PR = 3000

Taking ratio of both, R/2 = 1/30

R = 1/15 = 6.66%

Using options two and three,

Using statement 3, the simple interest for two years is Rs. 3000, so the simple interest for three years is Rs. 4500

Using statement 2, it implies that the principal is 19500-4500 = Rs. 15000.

Hence, the rate of interest is $$\frac{3000}{2*15000}=10\%$$

Hence, we can determine by taking any two statements at a time.

Using II, we know that there are 11k students in the class where k = 1, 2, 3,...
Using statement I, we know that 22 < 11k < 36 => Only possible value of 11k is 33
So, the question can be answered using I and II alone

As we know that P is the principal amount and rate of interest is R p.c.p.a., then:

From statement 1,

Difference = $$\frac{1}{10000} (PR^{2}) = 100$$

From statement 2,

$$\frac{(PR\times3)}{100}=19500-P$$

From statement 3,

$$\frac{PR \times 2}{100} = 3000$$

Dividing equations from statement 1 and statement 3

$$\frac{PR^2}{PR} = \frac{1000000}{150000} = \frac{20}{3}$$

R = $$\frac{20}{3}$$ p.c.p.a.

Here, the data in statement 1 is sufficient to answer the question.

Half an hour ago, the hour hand will be between 8 and 9. After half, the hour hand will be at 9 while minute hand will be on 12.

Therefore, option A is correct.

Here, we don't know principal and rate of interest. They form the two unknown variables.

Let Principal be P and Number of years be N.

II and III condition give two equations in terms of P and N.

I doesn't help with it's data and hence, it is redudant.

Correct option is Option C

Here, there are four unknown variables namely length of platform (X), time required to cross the platfom (n), length of train (y) and time required to cross signal pole (m).

With given information we can only form 3 equations

The number of variables (4) is more than the equations (3) that can formed.

Hence, we can say that the data is insufficient for calculation.

Therefore, E is the correct option.

Using statement 1, we get

$$P(1+\frac{R}{100})^2 - P - (P*R*2)/100 = 100$$

$$\frac{PR^2}{10000} = 100$$

$$PR^2 = 1000000$$

Using statement 3, we get

P*R*3/100 = 3000

PR = 100000

From these 2 statements we get R = 10%

Clearly, each statement alone is insufficient.

Let speed of the train = s kmph

I & II : Length of train = y, length of platform = x, time taken = n

=> $$\frac{x + y}{s}$$ = n

=> s = $$\frac{n}{x + y}$$ [ where, x,y,n are constants.]

Thus, I & II are sufficient.

II & III : Length of train = y, time taken = m

=> s = y/m

Thus, II & III are sufficient.

I & III : Let length of train = L

=> $$\frac{L}{s}$$ = m [From III]

& $$\frac{x + L}{s}$$ = n [From I]

Since, we have two variables 'L' & 's' and two equations. Therefore, we can find the speed of train.

Thus, I & III are sufficient.

Thus, any two of the three statements are sufficient.

Clearly, each statement alone is not sufficient.

And since we do not know the total no. of students in the class, we cannot determine the number of students passed in first class.

Thus, all statements combined together are not sufficient.

Both the statements I and II mean the same thing. ie A has 40% of the investment and B has 60% of the investment.

Statement III gives the value of one investment.

So, we can consider one of the two statements. With the ratio of investments and the value of one investment, we can calculate the other.

So, investment in B = 60* 45000/ 40 = 67,500

Clearly, each statement alone is insufficient as there is missing information about other persons.

I & II : Let A & B started with investment of 3 lakhs & 5 lakhs respectively. C joined after 6 months with 4 lakhs.

=> Ratio of profit = A : B : C = (3*12) : (4*6) : (5*12)

= 36 : 24 : 60 = 3 : 2 : 5

=> B will get two-tenth of the profit.

Thus, I & II together are sufficient.

After combining I & III or II & III, we will still have insufficient information.

Thus, ans - (a).

Using statement A alone, we cannot answer the question.

Using statement B alone also, we cannot answer the question. This is because, the statement talks about the 'number of students' but we cannot determine the number of students in course A.

Even by using both the statements, we cannot determine the number of students in course A of college X. So, option d) is the correct answer.

We cannot answer the question using either statement A or statement B alone. However, using both the statements, we can determine the answer as 83. Option e) is the correct answer.

S.P. = 100, we need to find the profit.

I : if S.P = 90, => profit % = 20%

=> $$\frac{90 - C.P.}{C.P.} * 100 = 20

=> 6 * C.P. = 450 => C.P. = 75

Therefore, Profit = 100 - 75 = Rs. 25

Thus, I alone is sufficient.

II : Let C.P. = 3x

=> Profit = 3x / 3 = x

=> 100 - 3x = x [Since, S.P. - C.P. = Profit]

=> x = 25 = Profit

Thus, II alone is sufficient.

Let digit at unit's place = y

Digit at ten's place = x

=> Number = 10x + y

Given : | x-y | = 6

I : y > x => y - x = 6

No. can be 17,28,39

Thus, I alone is not sufficient.

II : x + y = 12

case 1 : if x - y = 6

Solving the two equations, we get x = 9 & y = 3 => No. = 93

case 2 : if y - x = 6

We get, x = 3 & y = 9 => No. = 39

Thus, II alone is not sufficient.

I & II : y - x = 6 & x + y = 12

Solving, we get x = 3 & y = 9

=> number = 39

Thus, both I & II together are sufficient.

I : The difference between two digits is 9 only if the digits are 0 and 9.

No. using these digits = 90 and 09.

Since, 90 is the only two digit no. Ans - 90

Thus, I alone is sufficient.

II : Let one's digit be y and ten's digit be x. => Number = 10x + y

x + y = x - y

=> y = 0 & x can be any digit.

Thus, number can be 10,20,30,40...

Thus, II alone is not sufficient.

Simple Interest = PRT/100

Compound Interest = $$P * (1+R/n)^{nT}$$ - P where n is the compounding frequency

Statement I gives us 1 equation with 3 variables P, R and n. Hence, the answer cannot be found by Statement I alone.

Statement II gives 1 equation with P and R. Hence, the answer cannot be found by statement II alone.

Statement III gives us that n=1. Hence, the answer cannot be found using III alone.

Hence, all three statements are required to solve for P. Hence, option E.

Using statement 1, since a = 1
$$\frac{a^5b^3c^2}{a^4b^6c^2} = \frac{b^3c^2}{b^6c^2}$$
Using statement 2, since b = 1
$$\frac{a^5b^3c^2}{a^4b^6c^2} =\frac{a^5c^2}{a^4c^2}$$
If we combine both the statements,
then $$\frac{a^5b^3c^2}{a^4b^6c^2} = \frac{c^2}{c^2}$$
As it is given that c is a positive integer, we can cancel $$c^2$$ from numerator and denominator. Hence, the question can be answered using both the statements.

We cannot find the value of x using any one equation.
The given equations are 2x+3y = 12, 3x+8y = 24.
Here a/d = 2/3, b/e = 3/8 and c/f = 12/24 = 1/2.
Here a/d is not equal to b/e. Hence the value of x can be found.

From the statement 1, 4x+8y = 24
4(x+2y) = 4*6
x+2y = 6
Hence the statement can be answered using statement 1 alone.
From statement 2, 3x+8y = 9
We cannot find the value of x+2y using statement 2 alone.
Hence the correct answer is a.

Using statement 1,
MP = 140
Discount = 10%
SP = 90% of 140 = 126.
Hence the question can be answered using statement 1 alone.
Using statement 2,
Mark up% = 40% and discount% = 10%
We can only determine the profit% but we cannot determine the SP using statement 2 alone.
Hence the question can be answered using statement 1 alone.

Based on the first statement, we can derive the equation 2x +3y = 36.
Based on the second statement, we can derive the equation $$x^2 + y^2 = 10^2$$
By solving the both equations, we get two sets of values for x, y
X =6, Y = 8.

X = 5.07, Y = 8.61.

This impies X is less than Y.

Therefore it requires information from both the statements to solve the question.

I : No. of students can be = 21,22,23,24,25,26

Thus, I alone is not sufficient.

II : No. of students is multiple of 5 and is between 24 and 31

=> No. of students = 25,30

Thus, II alone is not sufficient.

I & II : From above two conclusions, only common number is 25

=> No. of students = 25

Thus, I & II together are sufficient.

I : More than four but less than ten students scored more than that of B. B’s rank is an odd number

=> B's rank can be 5 or 7 or 9.

Thus, I alone is insufficient.

II : The rank of C is 16th from the last, => C's rank from top = (25+1) - 16 = 10th.

B got more marks than C. Only two students are there between B and C.

=> B's rank from top = 7th

Thus 6 students scored more marks than B.

$$\therefore$$ II alone is sufficient.

=> Ans - (B)

I : Sum of digits of the number = 8

=> Number = 17,26,35,44...

$$\because$$ There is no unique answer. Thus, I alone is insufficient.

II : Sum of the digits is 3 more than the higher digit.

E.g. Number = 36 => Sum of digits = 9 and higher digit = 6

Similarly, numbers = 36,63,35,43...

Thus, II alone is insufficient.

I & II : Combining the two cases,

The possible numbers are = 35 , 53

Still we do not have any unique answer.

Thus, even I & II together are not sufficient.

From I : It is given that $$x \times y > 85$$

Let $$x = 100 , y = 100$$

=> $$x \times y = 100 \times 100 = 10000 > 85$$

and $$x + y = 100 + 100 = 200 > 85$$

Again let us assume $$x = 10 , y = 10$$

=> $$x \times y = 10 \times 10 = 100 > 85$$

But, $$x + y = 10 + 10 = 20 < 85$$

Thus, statement I alone is insufficient.

From II : We have no information of the second variable.

Thus, II alone is also insufficient.

From I & II : $$x \geq 84$$

and $$y \geq 2$$

=> $$x + y \geq 86$$, which answers our question.

Thus, I & II together are sufficient.

Considering statement 1 -

Soham did not have a two digit score so he must have correctly attempted all 10 questions i.e. score100 or wrongly attempted all 10 questions i.e. score 0 marks. But we know that he correctly attempted atleast 1 question. Hence his score must be 100.

Hence, statement 1 alone is sufficient to answer the question.

Considering statement 2 -

Suppose his present score is 70 (7 correct and 3 wrong). If he got 3 more questions wrong as compared to present, his new score will be 40 (4 correct and 6 wrong), which is less than 60.

Suppose his present score is 80 (8 correct and 2 wrong). If he got 3 more questions wrong as compared to present, his new score will be 50 (5 correct and 5 wrong), which is less than 60.

Hence, statement 2 alone is not sufficient to answer the question.

Hence option A is correct.

C.I. = $$P [(1 + \frac{R}{100})^T - 1]$$

I : Interest = 100 and sum = 1000 and time = 1 year

=> $$100 = 1000 [(1 + \frac{R}{100})^1 - 1]$$

=> $$\frac{100}{1000} = [(1 + \frac{R}{100}) - 1]$$

=> $$\frac{R}{100} = \frac{1}{10}$$

=> $$R = \frac{100}{10} = 10 \%$$

$$\therefore$$ C.I. after 3 years = $$1000 [(1 + \frac{10}{100})^3 - 1]$$

= $$1000 [(\frac{11}{10})^3 - 1]$$

= $$1000 [\frac{1331}{1000} - 1] = 1000 \times \frac{331}{1000}$$

= $$331$$

Thus, I alone is sufficient.

II : Sum = 1000 , rate = $$R \%$$

S.I. after 2 years = $$\frac{P \times R \times T}{100}$$

= $$\frac{1000 \times R \times 2}{100} = 20R$$

C.I. after 1st year = $$\frac{R}{100} \times 1000 = 10R$$

C.I. after 2nd year = $$10R + \frac{R}{100} \times 10R = 10R + \frac{R^2}{10}$$

=> Required difference = 10

=> $$(10R + 10R + \frac{R^2}{10}) - (20R) = 10$$

=> $$\frac{R^2}{10} = 10$$

=> $$R^2 = 10 \times 10 = 100$$

=> $$R = \sqrt{100} = 10 \%$$

$$\therefore$$ C.I. after 3 years = $$1000 [(1 + \frac{10}{100})^3 - 1]$$

= $$1000 [(\frac{11}{10})^3 - 1]$$

= $$1000 [\frac{1331}{1000} - 1] = 1000 \times \frac{331}{1000}$$

= $$331$$

Thus, II alone is sufficient.

Thus, either statement alone is sufficient.

Clearly, we have to use both equations.

Let distance between A and B = $$d$$ km

Let speed of boat in still water = $$x$$ kmph

and speed of current = $$y$$ kmph

Using, $$time = \frac{distance}{speed}$$

=> $$\frac{d}{x + y} = 2$$ ---------Eqn(1)

and $$\frac{d}{x - y} = 4$$ ---------Eqn(2)

Dividing equation (2) by (1)

=> $$\frac{x + y}{x - y} = \frac{4}{2}$$

=> $$x + y = 2x - 2y$$

=> $$2x - x = y + 2y$$

=> $$x = 3y$$

=> $$\frac{x}{y} = \frac{3}{1}$$

$$\therefore$$ Ratio between speed of boat and current is known and not the exact value.

Thus, even both statements together are insufficient.

I : We know that ‘$’ means ‘÷’, ‘*’ means ‘x’ and ‘#’ means ‘+’

=> 144 $ 16 *7 # 9 = $$144 \div 16 \times 7 + 9$$

= $$63 + 9 = 72$$

Thus, statement I alone is sufficient.

II : 16 $ 4 *2 # 2 =10

=> $$16 \div 4 \times 4 + 2 = 10$$

=> ‘$’ means ‘÷’, ‘*’ means ‘x’ and ‘#’ means ‘+’

=> 144 $ 16 *7 # 9 = $$144 \div 16 \times 7 + 9$$

= $$63 + 9 = 72$$

Thus, II alone is sufficient.

$$\therefore$$ Either statement alone is sufficient.

Let number of oranges be $$o$$, apples be $$a$$ and bananas be $$b$$.

I : Total fruits = 100

Number of oranges and bananas = $$o+b=\frac{50}{100}\times100=50$$

=> $$a=100-50=50$$

Thus, I alone is sufficient.

II : We don't know the total fruits, hence II alone is insufficient.

$$\therefore$$ I alone is sufficient.

=> Ans - (A)

Brinda's rank from top = 17th

I : There are 70 students in her class.

=> Brinda's rank from last = $$(70+1)-17=54$$

Thus, I alone is sufficient.

II : Nisha's rank from top = 20th and from last = 51st

Thus, total students in class = $$(51+20)-1=70$$

=> Brinda's rank from last = $$(70+1)-17=54$$

Thus, II alone is sufficient.

$$\therefore$$ Each statement alone is sufficient.

=> Ans - (C)

I : Since, we do not know the principal amount, thus I alone is not sufficient to find the rate of interest.

II : Let principal be $$100x$$

=> Amount after 5 years = $$\frac{180}{100} * 100x = 180x$$

=> interest after 5 years = $$180x - 100x = 80x$$

$$\because$$ $$r = \frac{S.I. \times 100}{P \times T}$$

=> $$r = \frac{80x \times 100}{100x \times 5}$$

=> $$r = 16$$%

Thus, II alone is sufficient.

I : Let the number of boys and girls be $$8x$$ and $$5x$$ resp

=> $$8x - 5x = 24$$

=> $$x = \frac{24}{3} = 8$$

Hence, total number of children in the class = $$13x$$ = 13 * 8 = 104

Statement I alone is sufficient

Statement II does not give any relation between the number of boys and girls. Thus, it is not sufficient.

Let the number be $$10x + y$$

I : $$x - y = 7$$ or $$y - x = 7$$

II : $$x + y = 11$$

I & II : Solving $$x - y = 7$$ and $$x + y = 11$$

=> $$x = 9 , y = 2$$

Solving $$y - x = 7$$ and $$x + y = 11$$

=> $$x = 2 , y = 9$$

Hence, the possible numbers are 92 and 29. Since, unique answer is not determined

Ans - (D)

I : Speed = 12 kmph

= 12 * $$\frac{5}{18} = \frac{10}{3}$$ m/s

II : Circumference of ground = $$\frac{22}{7} * 700$$ = 2200 m

When both statements are combined, the required time will be

Time = $$\frac{Distance}{Speed}$$

= $$\frac{2200 * 3}{10}$$ = 660 s

From statement II and I
We can conclude that he had 1, 2, 4 and 8 Kgs as the weight blocks.

Satement III is redundant, it serves no purpose to solve the question.

Hence, option B is the correct answer.

From statement 1, we can infer that 45 = 10Cx. A bit of trial and error will give 45 = 10C2. But, 10C8 = 10C2. We know that 2 students have either been selected or rejected.

From statement 2, we know that the number of people selected is less than the number of people rejected.
Combining statements 1 and 2, we can determine that 2 people must have been selected for the competition. Hence, option D is the right answer.

We cannot answer the question using I as Mohit can be 2nd or 13.

Using only 2, we can determine Mohit's rank.

Using 3, we can't determine Mohit's rank as we do not know the total number of participants.

From statement 1 alone we can find that Mayank took 300/50= 6 days to complete the book. But we can’t tell till Wednesday how many pages did he read.
From statement 2 alone we find that Mayank reads last 75 pages on Thursday, so till Wednesday he must’ve read 300 - 75 = 225 pages irrespective on which day he started reading the book. Hence statement 2 alone is sufficient to answer the question.
Hence option B is correct.

I : Harish missed his usual train at 10.25 a.m. and in every 5 minutes, a train comes.

Harish could have caught any train at 10.30 a.m. , 10.35 a.m. , 10.40 a.m. and so on.

There is no information about that, so it is not possible to determine which train Harish caught.

Thus, I alone is insufficient.

Even in statement II or even after combining both the statements, the same problem arises.

Thus, data in both statements even together is not sufficient to answer the question.

Let rate of interest = $$R \%$$

I : P = Rs. 25,000 and time period = 2 years

=> $$S.I. = \frac{P \times R \times T}{100}$$

= $$\frac{25000 \times R \times 2}{100} = 500 R$$

Interest obtained from compound interest = $$P [(1 + \frac{R}{100})^T - 1]$$

=> $$(500R + 250) = 25000 [(1 + \frac{R}{100})^2 - 1]$$

=> $$250 (2R + 1) = 25000 [(1 + \frac{R}{100})^2 - 1]$$

=> $$2R + 1 = 100 (1 + \frac{R^2}{100^2} + \frac{2 R}{100} - 1)$$

=> $$2R + 1 = \frac{R^2}{100} + 2R$$

=> $$R^2 = 100 \times 1 = 100$$

=> $$R = \sqrt{100} = 10 \%$$

Thus, statement I alone is sufficient.

II : Let original sum = simple interest = $$Rs. P$$ and time period = 10 years

=> $$S.I. = \frac{P \times R \times T}{100}$$

=> $$P = \frac{P \times R \times 10}{100}$$

=> $$R = \frac{100}{10} = 10 \%$$

Thus, statement II alone is also sufficient.

Thus, either statement alone is sufficient.

To find the number of plants planted, we need to find the number of columns and rows and then multiply them.

Let number of columns = $$c$$ and number of rows = $$r$$

(I) : Number of columns is 4 more than rows

=> $$c - r = 4$$

But, we do not know the exact value. So, statement I alone is not sufficient.

(II) : In each column ,number of trees are even, => number of rows are even in number

Again, we do not know the exact value. So, statement II alone is not sufficient.

Combining above statements, we still do not know the exact number of columns and rows. So, statement I and II together are not sufficient.

=> Ans - (D)

I : Let present age of son = $$x$$ years

=> Present age of father = $$5x$$ years.

There is no other information, so statement I is not sufficient.

Similarly, II alone is not sufficient.

Combining both statements, we get :

=> $$(5x - 5) = 15 (x - 5)$$

=> $$5x - 5 = 15x - 75$$

=> $$15x - 5x = 75 - 5$$

=> $$10x = 70$$ => $$x = \frac{70}{10} = 7$$

$$\therefore$$ father's age = $$5 \times 7 = 35$$ years.

Thus, both statements together are sufficient.

Using statements I and II together, we can exactly determine the amount spent my Manoj on entertainments. Thus, III becomes redundant.
Also, using statements II and III together, we can exactly determine the amount spent my Manoj on entertainments. Thus, I becomes redundant.
Hence, either I or III can become redundant.
Hence, option C is the right choice.

Using statement I alone:
Let 'a' be the HCF of x and y. Then, LCM = 20x
Hence, the product of the numbers = x*20x = 20x$$^2$$. We can see that the product depends on x. Hence, we can't find the value of x*y using statement I alone.

Using statement II alone:
We have no information regarding LCM of x and y. Hence, using statement II alone we can't find out the value of x*y.

Using both the statements together:
From statement (I) we can see that the value of x*y = 20x$$^2$$
From statement (II) we can see that x = 30. Hence, we can find out the value of 20x$$^2$$. Therefore, option C is the correct answer.