SNAP Data Sufficiency Questions

I) $$a-c$$ is negative as $$a$$ is negative and $$c$$ is positive. Thus, the value of $$a-c$$ will be smaller than $$a$$(since more negative is added to a negative number). $$a-b$$ can be negative or positive depending upon the value of $$a$$ and $$b$$, but it can be concluded that this number will be greater than $$a$$(since a positive number is added to a negative number). Thus, $$(a-c)<(a-b)$$. Thus, the statement I is false.

II) Given $$a<b$$, $$\frac{a}{c}<\frac{b}{c}$$ as $$c$$ is positive and $$a < b$$. Therefore, statement II is true

III) $$a$$ is negative and $$b$$ is negative, this implies $$\frac{a}{b}$$ is positive. $$a$$ is negative and $$c$$ is positive, this implies $$\frac{a}{c}$$ is negative. Therefore, $$\frac{a}{b}>\frac{a}{c}$$. Statement III is true

Answer is option D

Using $$a_n$$ = a+(n-1)d

=> $$a_57$$=a+56d

From 1, the value of d can be obtained.

From 2, an equation in a and d is obtained. 

Hence, both 1 and 2 are necessary to calculate the value of $$a_57$$.

If $$\ \frac{\ a}{\ \ b}$$ is less than 1, then $$\ \frac{\ a+7}{\ \ b+7}$$ > $$\ \frac{\ a}{\ \ b}$$ given that a,b>0

Hence, if I is given. II or III are redundant.

Considering statement 1 alone, we can make a table as follow:

We get S-1220=P and S'-1400=1.32P

Since there are 3 variables, S, S' and P, we cannot get the values of the variable using the two equations alone. 

So, Statement 1 alone is not sufficient.

Using statement 2, 1.32P= 4300

=> P= 4300/1.32. But again, two variables, S and S' are unknown.

So, So, Statement 1 alone is not sufficient.

When we use both the given statements, we will have 

S-1220=4300/1.32 and 

S'-1400=4300.

Using these two equations, we can find S and S' respectively and hence answer the question asked.

Therefore, both statements are necessary to answer the question.

Understanding the premise of the question, we can infer that the price of the laptop when taken individually is, let's say x and that of the printer is y. 

But, when taken together under the bundle offer, the price is lesser than (x+y), because of the discount given, let's say a%.

Statement 1 gives us the discounted price of the printer and laptop taken together, i.e. $$\ \left(x+y\right)\left(1-\ \frac{\ a}{100}\right)$$

We have three variables and cannot solve this equation.

Statement 2 gives us the value of x. But still, two variables, a and y are unknown.

Even if we use both the statements together, we will have 2 variables and only 1 equation.

Therefore, both statements are insufficient to find the value of y.

Considering statement 1 alone,

Let the number of balls in bag III be x.

The number of balls in bag I becomes x+5.

And the rest, which is 10- (x+x+5)= 5-2x balls are with C.

We cannot find the value of x, so Statement 1 alone is not sufficient.

Considering statement 2 alone, 

Let the number of balls in bag I be 2a

Balls in bag II will be a and the rest, i.e 10-3a will be in bag 3.

Again, we cannot find the value of a in this case, and hence, Statement 2 alone is not sufficient.

Considering both the statements and using the number of balls with each one of them as found using statement 1, we get

$$5+a=2\left(5-2a\right)$$

=>5+a= 10-4a

=> 5a=5, or a=1.

So, balls in Bag I= 5+a=6

balls in Bag II= 5-2a=3

balls in Bag III= a= 5.

Therefore, both statement I and II are required.

Let the temperature on Monday, Tuesday, Wednesday and Thursday be a,b,c and d respectively

It is given that a+b+c = 3*37 = 111...(I)

              and b+c+d = 3*34 = 102 ... (II)

Using statement (I) alone we get $$d\ =\ \ \frac{\ 4}{5}a$$. Which can be put in eqn (I), after which we can obtain the value of d

Using statement (II) alone we can add eqn(I) and eqn(II) to get value of b+c and it can be used to get value of d from eqn(II)

Using statement (III) alone we a-d = 9 which can be obtained by eqn(I) and (II) and n hence no new information is given. Statement (III) is no sufficient to get the required answer.

Statement 1: Let the smaller number be x, then the bigger number becomes x + 6.

Statement 2: Let the smaller number be a and the bigger number be b. Given that 40% of a is 30% of b,

$$\frac{40}{100}\ \times\ a\ =\ \frac{30}{100}\ \times\ b$$

$$\frac{a}{b}\ =\ \frac{3}{4}\ $$

Statement 3: Let the smaller number be a and the bigger number be b. The ratio of half of the bigger number to one-third of the smaller number is 2: 1, which means,

$$\dfrac{\frac{b}{2}}{\frac{a}{3}}\ =\ \frac{2}{1}$$

$$\frac{3b}{2a}\ =\ \frac{2}{1}$$

$$\frac{a}{b}\ =\ \frac{3}{4}$$

Case 1 : I and II

Let smaller number be  x => bigger number = x+6

$$\frac{40}{100}\cdot x=\frac{30}{100}\left(x+6\right)$$

=> x=18 , Sum of bigger and smaller number = 42

Case 2 : I and III

Let smaller number be x => bigger number = x+6

$$\frac{\left(x+6\right)}{2}\cdot\frac{3}{x}=\frac{2}{1}$$

On solving, x=18. So sum = 42

Case 3 : II and III

The conclusion from both statements is the same, and the sum of the numbers cannot be obtained using both statements.

Hence, the correct answer is option C.

Let total work to be done is 48 units

Case 1 : I and II

A takes 16 hours => B=24 hours  (since A is 50 percent more efficient than B)

A does 3 units/hour

B does 2 units/hour

Total per day work = 5 units

Total time taken = 48/5 hr

Case 2 : I and III

B takes 24 hours => A = 16 hours (since A is 50 percent more efficient than B)

Together they take = 48/5 hours

Case 3 : II and III 

B takes 24 hours

A takes 16 hours

Together they take = 48/5 hours

A=$$P\left(1+\frac{TR}{100}\right)$$

From 2 and 3 , A= 750 , T=5 and R=8

P can be found out by substituting these values in the above equation.