10 Most Common 7th Grade IAR Math Questions

B
Write the numbers in order:
\(2, 19, 28, 28, 35, 44, 67\)
Since we have \(7\) numbers (\(7\) is odd), then the median is the number in the middle, which is \(28\).

C
Three times \(24,000\) is \(72,000\). One-sixth of them canceled their tickets.
One sixth of \(72,000\) equals \(12,000\) \((\frac{1}{6}) × 72,000 = 12,000\).
\(60,000\) \((72,000 – 12,000 = 60,000)\) fans are attending this week

A
It needed to have a ratio to find the value of \(x\).
\(\frac{45}{40}=\frac{2x+4}{16}⇒ 40(2x+4)=45×16 ⇒ x=7\)

B
\(8 (4-(-8))=96\)

A
\(\frac{1}{3}x + \frac{1}{6}x + \frac{1}{4}x + 12= x\)
\((\frac{1}{3} + \frac{1}{6} + \frac{1}{4}) x+ 12= x\)
\((\frac{9}{12})x+ 12 = x\)
\(x = 48\)
In a bag of small balls \(\frac{1}{6}\) are white then: \(\frac{48}{6} = 8\)

B
Use the information provided in the question to draw the shape.
Use Pythagorean Theorem: \(a^2 + b^2 = c^2\)
\(40^2 + 30^2 = c^2 ⇒ 1,600 + 900 = c^2 ⇒ 2,500 = c^2 ⇒ c = 50\)

B
The question is this: \(530.40\) is what percent of \(624\)?
Use the percent formula:
\(part = \frac{percent}{100}× whole\)
\(530.40= \frac{percent}{100}× 624 ⇒ 530.40 = \frac{percent ×624}{100}⇒53,040 = percent ×624\)
\(⇒percent = \frac{53,040}{624}= 85\)
\(530.40\) is \(85 \%\) of \(624\). Therefore, the discount is: \(100\% – 85\% = 15\%\)

B
If the score of Mia was \(60\), therefore the score of Ava is \(30\). Since the score of Emma was half as that of Ava, therefore, the score of Emma is \(15\).

D
If \(17\) balls are removed from the bag at random, there will be one ball in the bag.
The probability of choosing a brown ball is \(1\) out of \(18\). Therefore, the probability of not choosing a brown ball is \(17\) out of \(18\) and the probability of having not a brown ball after removing \(17\) balls is the same.

C
The weight of \(12.2\) meters of this rope is: \(12.2 × 600 \space g = 7,320 \space g\)
\(1\space kg = 1,000 \space g\) therefore,
\( 7,320 \space g ÷ 1,000 = 7.32 \space kg\)